Class 09 | Ganita Manjari 01 | Ch. 04 Exploring Algebraic Identities

Class 9 | Ganita Manjari | Ch. 04  Exploring Algebraic Identities

Page 68

Example 1: Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. 

Thus, 1 + 9 = 10. 

Then subtract twice the middle square from this sum. 

This leads to 

10 – (2×4) 

= 10 – 8 

= 2.

Now try the same process with another set of three consecutive square numbers. 

9, 16, 25.

9 + 25 = 34

= 34 – (2x16) 

= 34 – 32

= 2

The Result is always 2

For example, consider the consecutive squares 25, 36, 49.

a², b², c²

(a² + c²) – (2 × b²) = 2

Applying the same rule we get 

(25 + 49) – (2×36) 

= 74 – 72 

= 2.

Page 69

Think and Reflect

Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.

Pattern with 4 Consecutive Square Numbers

A², B², C², D²

(A² + D²) – (B² + C²) = 4

Take four consecutive square numbers:

[ 1, 4, 9,16 ]

Add the first and fourth squares:

[ 1 + 16 = 17 ]

Add the second and third squares:

[ 4+9=13 ]

Now find the difference:

[ 17-13=4 ]

Try another set:

[ 4, 9, 16, 25 ] 

= [ (4+25) – (9+16) ] 

= [ 29 – 25]

={4}

So, we can observe the pattern:

[{First square + Fourth square − Second square − Third square = 4}} ]

This pattern works for any four consecutive square numbers.

VISUALISING IDENTITIES

In this section we will revisit some algebraic identities that we have studied in earlier grades and try to visualise them using geometrical models. In particular, we will use squares and rectangles to represent terms.

Consider two line segments of lengths a and b units, respectively, and make a longer line segment of length (a + b) units as shown in Figure. 

We can now construct a square of side (a + b) units and partition it into smaller squares and rectangles as shown in Fig. 4.2.

Figure: Square of side (a + b) units

Observe that the area of the outer square is (a + b)² The area of the larger square inside the outer square is a² while the area of the smaller square is b².

The areas of the two rectangles are ab each. Together they make the bigger square; hence we can conclude that

(a + b)² = a² + 2ab + b²

larger square inside the outer square is a² while the area of the smaller square is b² The areas of the two rectangles are ab each. 

Together they make the bigger square; hence we can conclude that (a + b)² = a² + 2ab + b²

From Fig. 4.2 it is clear that (a + b)² = a² + 2ab + b² for all a and b when a and b are lengths of line segments.

Page 70

Example 2: 

Let a = – 2 and b = – 3

(a + b)² = a² + 2ab + b²

Then 

[(–2) + (–3)]² = (–2)² + 2×(–2)(–3) + (3)²

            [(–5)]² = 4 + 12 + 9

                   25 = 25

So, a² + 2ab + b² = (a + b)² seems to be true for rational numbers too. 

Finding 

(a + b)² = a² + 2ab + b²

But we are still not sure if it is true for all numbers. To verify this, let us investigate further using the distributive property of numbers:

(a + b)² 

= (a + b)(a + b) 

= a(a + b) + b(a + b)

= a² + ab + ba + b² 

= a² + 2ab + b² . 

Recall that in Grade 8 you were introduced to 

(a + b)² = a² + 2ab + b² as an identity.

What is the difference between an equation and an identity?

An algebraic identity is an equation that is true for all values of the variables occurring in it, while an equation need not be true for all values.

For example, x² – 1 = 24 is true for only x = 5 or –5. 

Hence, it is an equation. 

But (x + y)² = x² + 2xy + y² is true for all values of x and y. 

Therefore, this equation identity.

By now you must have observed that 

(a+b)² ≠ a² + b² . 

But can you find out which one of them will be greater

For example, if a = 10 and b = 2 what are the values of (a + b)² and a² + b² ?

(a + b)² = (10 + 2)² = 12² = 144 

and a² + b² = 10² + 2² = 100 + 4 = 104

So in this case, 

(a + b)² > a² + b²

But is it true for all numbers a and b?

Page 71

Think and Reflect

1. What can you say about a and b if 

(a + b)² < a² + b² ?

Sol. When it

(a+b)² = a² + b² ?

Using the expansion:

a² + 2ab + b² < a² + b²

Subtract from both sides:

2ab < 0

ab < 0

This means a and b have opposite signs.

Ans.

when one of a and b is positive and the other is negative.

2. What can you say about a and b if 

(a + b)² > a² + b² ?

Sol. When is

(a+b)² > a² + 2ab + b² ?

Using the expansion:

a² + 2ab + b² > a² + b²

2ab > 0

ab > 0

This means and have the same sign.

Ans.

when both and are positive or both are negative.

3. When will (a + b)² be equal to a² + b²

Sol. When is

(a+b)² = a² + b²?

Using the expansion:

a² + 2ab + b² = a² + b²

2ab = 0

ab = 0

So either or .

Ans. 

when one of the numbers is zero.

Did you observe that (a + b)² and a² + b² are both positive? What term will decide which is larger? Use the expansion of (a + b)² to decide.

Sol.

Conclusion

Both a and b are positive quantities.The term that decides which is larger is:

The middle term determines whether is greater than, smaller than, or equal to .

We can use the identity (a + b)² = a² + 2ab + b² to find the square of general binomial expressions.

We know that

(a+b)² = a² + 2ab + b²

So, the difference between and depends on the term .

Example 3: Let us try to expand (5x + 2y)² 

Sol.

Here a = 5x and b = 2y 

Then 

(5x + 2y)² 

= (5x)² + 2(5x)(2y) + (2y)² 

= 25x² + 20xy + 4y²

We can also use the identity to help us with numerical calculations.

Example 4: To calculate 43².

Sol.

we can write it as 

= 43²

= (40 + 3)² 

= 40² + 2×40×3 + 3² 

= 1600 + 240 + 9 

= 1849

Page 71

EXERCISE SET 4.1

1. Using the identity (a + b)² = a² + 2ab + b² expand the following:

(i) (7x + 4y)²

(ii) [(7/5) x + (3/2) y]² 

(iii) (2.5p + 1.5q)²

(iv) [(3/4)s + 8t]²

(v) [(x) + 1/(2y)]²

(vi) [(1/x) + (1/y)]²

Sol.

(i) (7x + 4y)²

= (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

(ii) [(7/5)x + (3/2y]²

= [(7/25)x]² + 2(7/5x)(3/2y) + [(3/2)y]²

= 49/4 x² + 21/5 xy + 9/4 y²

Or

(iii) (2.5p + 1.5q)²

= (2.5p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.25p² + 7.5pq + 2.25q²

(iv) [(3/4)s + 8t]²

= [(3/4)s]² + 2.(3/4)s.(8t) + (8t)²

= 9/16 s² + 12st + 64t²

Or

(v) [(x) + 1/(2y)]²

= (x)² + 2(x)(1/2y) + [(1/2y)]²

= x² + x/y + 1/4y²

Or

(vi) [(1/x) + (1/y)]²

= (1/x)² + 2(1/x)(1/y) + (1/y)²

= 1/x² + 2/x.y + 1/y²

Or

2. Using the same identity, find the values of the following:

(i) (64)²

(ii) (105)²

(iii) (205)²

Sol.

we can write it as 

(i). (64)²

= (60 + 4)² 

= 60² + 2×60×4 + 4² 

= 3600 + 480 + 16 

= 4096

(ii). (105)²

= (100 + 5)² 

= 100² + 2×100×5 + 5² 

= 10000 + 1000 + 25

= 11025

(ii). (205)²

= (200 + 5)² 

= 200² + 2×200×5 + 5² 

= 40000 + 2000 + 25

= 42025

4.3 FACTORISATION OF ALGEBRAIC EXPRESSIONS USING IDENTITIES

The identity (a + b)² = a² + 2ab + b² can also be used to find factors of some algebraic expressions.

Page 72

Example 5: Consider the algebraic expression x² + 4x + 4 

Sol.

We observe that 

x² = (x)² 4 = 2² and 4x = 2(x)(2) 

Hence, 

x² + 4x + 4 

= x² + 2(x)(2) + 2² can be compared with 

= a² + 2(a)(b) + b² 

where and a x, b = 2

We conclude that x² + 4x + 4 

= (x + 2)² 

Therefore, we can say that (x – 2) is a factor of x² + 4x + 4

Example 6: Let us try to find factors of another algebraic expression: 

36x² + 12x + 1

Sol.

Writing this in the form 

a² + 2ab + b² we get 

36x² + 12x + 1 

= (6x)² + 2(6x)(1) + 1² 

where a = 6x and b = 1 

Therefore, 36x² + 12x + 1 

= (6x + 1)²

Thus (6x + 1) is a factor of 36x² + 12x + 1

Example 7: Let us try to factor 

50p² + 60pq + 18q² What will a and b be in this case?

Sol.

Try to think of a term whose square is 50p² It is sqrt(50) * p But if we want to avoid using the square root symbol we may proceed as follows:

We observe that 2 is a common factor of the terms 50p ^ 2 60pq, 18q². 

Thus 

50p² + 60pq + 18q² 

= 2(25p² + 30pq + 9q²)

after taking 2 as a common

= 2[(5p)² + 2(5p)(3q) + (3q)²] 

= 2 [(5p + 3q)]² 

= 2 (5p + 3q)²

Page 73

Think and Reflect

Q. What if we replace 'b' by '–b' in (a + b)² = a² + 2ab + b² ?

Sol. 

[(a) + (–b)]² 

= (a)² + 2(a)(–b) + (–b)²

We get 

(a – b)² = a² – 2ab + b² 

which is also an identity and can be used in ways similar to (a + b)² = a² + 2ab + b²

Note 1. Any three consecutive numbers can be taken as (n –1), n and (n + 1) .

Their respective squares are of the form 

(n –1)², n² and (n + 1)²

Sol. 

The sum of the smallest and largest squares is (n – 1)² + (n + 1)² 

= n² – 2n + 1 + n² + 2n + 1 

= 2n² + 2

 This is in fact a proof of the fact that 

If we add the smallest and largest of any three consecutive square numbers and subtract two times the middle square number, we will always arrive at 2.

Subtracting 2n² from this leads to 2. Hence you always get 2!

(smallest²) + (largest²) – (two × middle²) = 2 

Just as the identity 

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

Example 8: Calculate 29² Or Find the square of 29

Sol.

We can express 29² as 

= (30 – 1)² 

= 30² – 2 × 30 × 1 + 1² 

= 900 – 60 + 1 

= 840 + 1 

= 841

To visualise the identity 

(a – b)² = a² – 2ab + b²

let us draw a square of side a units and split a in two parts, one of length (ab) units and another of length b units. Then the figure will look like this.

The area of the big square is a² square units. The small square has an area of (a – b)² square units. 

The larger rectangle has area ab square units while the smaller rectangle has area b (a – b) square units. Thus, to obtain (a – b)² we can subtract the areas of the rectangles from the big square.

We obtain 

Finding 

(a – b)² = a² – 2ab + b²

(a – b)² 

= a² – ab – b(a – b) 

= a² – ab – ba + b² 

= a² – 2ab + b² 

Page 74

EXERCISE SET 4.2

1. Factor completely:

(i) 9x² + 24xy + 16y²

(ii) 4s² + 20st + 25t²

(iii) 49x² + 28xy + 4y²

(iv) 64p² + (32/3) × pq + (4/9) × q²

*(v) 3a² + 4ab + (4/3) × b²

*(vi) (9/5) s² + 6sv + 5v²

Sol.

(i) 9x² + 24xy + 16y²

= 9x² + 12xy + 12xy + 16y²

= 3x(3x + 4y) + 4y(3x + 4y)

= (3x + 4y)(3x + 4y)

= (3x + 4y)²

Or

(i) 9x² + 24xy + 16y²

9x² = (3x)²  and  16y² = (4y)²

Middle term: 

= 2(3x)(4y) 

= 24xy 

So, 

9x² + 24xy + 16y ² = (3x + 4y)²

(ii) 4s² + 20st + 25t²

= 4x² + 10xy + 10xy + 25y²

= 2x(2x + 5y) + 5y(2x + 5y)

= (2x + 5y)(2x + 5y)

= (2x + 5y)²

(iii) 49x² + 28xy + 4y²

= 49x² + 14xy + 14xy + 4y²

= 7x(7x + 2y) + 2y(7x + 2y)

= (7x + 2y)(7x + 2y)

= (7x + 2y)²

(iv) 64p² + (32/3) × pq + (4/9) × q²

= (8p)² + 2 × 8p × ⅔q  + (⅔q)²

= (8p + ⅔q)²          [by a² +2ab + b² = a² – b²]

(v) 3a² + 4ab + (4/3) × b²

= (√3p)² + 2 × √3p × 2/√3 q  + (2/√3.q)²

= (√3p + 2/√3.q)²      [by a² +2ab + b² = a² – b²]

*(vi) (9/5) s² + 6sv + 5v²

= (3/√5. s)² + 2 × 3√5s × √5v  + (√5.v)²

= (3√5 s + √5 v)²      [by a² +2ab + b² = a² – b²]

2. Find the values of the following using the identity (a – b)² = a² – 2ab + b²

(i) (79)²

(ii) (193)²

(iii) (299)²

Sol.

(i) 79²  

= (80 – 1)² 

= 80² – 2 × 80 × 1 + 1² 

= 6400 – 160 + 1 

= 6240 + 1

= 6241

(ii) 193²  

= (200 – 7)² 

= (200)² – 2 × 200 × 7 + 7² 

= 40000 – 2800 + 49

= 37200 + 49

= 37249

(iii) 299²  

= (300 – 1)² 

= (300)² – 2 × 300 × 1 + 1² 

= 90000 – 600 + 1

= 89400 + 1

= 89401

4.4. MORE IDENTITIES

Finding 

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

What will happen if we want to find the square of the sum of three numbers a, b and c, 

that is, 

(a + b + c)²

Let us replace b + c by d.

We already know, (a + d)² = a² + 2ad + d² .

Thus, replacing d by (b + c), we get 

a² + 2ad + d²  

= a² + 2a(b + c) + (b + c)² 

= a² + 2ab + 2ac + b² + 2bc + c² .

= a² + b² + c² + 2ab + 2bc + 2ca

So, 

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

It may be more convenient to remember this as.

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

Let us interpret this geometrically by drawing a square of side a + b + c as shown in Figure 

Page 76

Think and Reflect

Label the squares and rectangles in Fig. 4.4 so that it represents the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

To represent the identity

[ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.]

Fig. 4.4 में squares और rectangles को इस प्रकार label करें:

3 squares: (a², b², c²)

2 rectangles: (ab, ab)

2 rectangles: (bc, bc)

2 rectangles: (ca, ca)

अर्थात चित्र में कुल 9 भाग होंगे:
[ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.]

या

[ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.]

Example 9: Let us use this identity to find the square of a number, say 119:

119² 

= (100 + 10 + 9)²

= 100² + 10² + 9² + 2(100)(10) + 2(100)(9) + 2(10)(9)

= 10000 + 100 + 81 + 2000 + 1800 + 180 

= 14161 .

So far we have verified the following three identities and used them for performing calculations and manipulating algebraic expressions:

1. (a + b)² = a² + 2ab + b²

2. (a – b)² = a² – 2ab + b²

3. (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

Page 76

EXERCISE SET 4.3

1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) 117²

(ii) 78²

(iii) 198²

(iv) 214²

(v) 1104²

(vi) 1120²

Sol.

(i) 117²
Sol.
117² = (100 + 10 + 7)²
= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 10000 +100 + 49 + 2000 + 140 + 1400 = 13689

(ii) 78²
Sol.
78² = (80 – 2)² = 80² – 2 × 80 × 2 + 2²
[∵(a – b)² = a² – 2ab + b²]
= 6400 – 320 + 4
= 6084

(iii) 198²
Sol.
198² = (200 – 2)²
= 200² + 2² – 2 × 200 × 2
[∵(a – b)² = a² – 2ab + b²]
= 40000 – 800 + 4
= 39204

(iv) 214²
Sol.
214² = (200 + 10 + 4)²
= 200² + 10² + 4² + 2(200)(10) + 2(10)(4) + 2(4)(200)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 40000 + 100 + 16 + 4000 + 80 + 1600
= 45796

(v) 1104²
Sol.
1104² = (1000+ 100 + 4)²
= 1000² + 100² + 4² + 2(1000) (100) + 2(100)(4) + 2(4)(1000)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816

(vi) 1120²
Sol.
1120² = (1000 + 100 + 20)²
= 1000² + 100² + 20² + 2(1000)(100) + 2(100)(20) + 2(20)(1000)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= 1000000 + 10000 + 400 + 200000 + 4000 + 40000
= 1254400

2. Factor using suitable identities:

(i) 16y² – 24y + 9

(ii) (9/4) s² + 6st + 4t²

(iii) (m²)/9 + (mk)/3 + (k²)/4 + 3nk + 2mn + 9n²

(iv) (p²)/16 – 2 + 16/(p²)

(v) 9a² + 4b² + c² – 12ab + 6ac – 4bc

Sol.

Question 2.
Factor using suitable identities:
(i) 16y² – 24y + 9
Solution:
16y² – 24y + 9 = (4y)² – 2 × (4y) × 3 + (3)²
= (4y – 3)²
[∵(a – b)² = a² – 2ab + b²]

(ii) 94s² + 6st + 4t
Solution:
94s² + 6st + 4t
= (32s)² + 2 × (32s) × (2t) + (2t)²
= (32s + 2t)²

(iii) m29+mk3+k24 + 3nk + 2mn + 9n²
Solution:
= (m3)2+(k2)2 + (3n)² + 2 × (m3) × (3n) + 2 × (k2)×(m3) + 2 × (k2) × (3n)
= (m3+k2+3n)2
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

(iv) p/216 – 2 + 16/p2
Solution:
(p4)2 – 2 × p4×4p+(4p)2
= (p4−4p)2
[∵(a – b)² = a² – 2ab + b²]

(v) 9a² + 4b² + c² – 12ab + 6ac – 46c
Solution:
9a² + 4b² + c² – 12ab + 6ac – 46c
= (3a)² + (-2b)² + (c)² + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)
= [3a + (-2b) + c]²
= (3a – 2b + c)²

3. Expand the following using the identity

(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

(i) (p + 3q + 7r)²

(ii) (3x - 2y + 4z)²

Sol.

Question 3.
Expand the following using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)²
Solution:
(p + 3q + 7r)²
= p² + (3q)² + (7r)² + 2{p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42 qr + 14pr

(ii) (3x – 2y + 4z)²
Solution:
(3x – 2y + 4z)²
= (3x)² + (-2y)² + (4z)² + 2(3x)(-2y) + 2(-2y) + 2(-2y)(4z) + 2(3x)(4z)
= 9x² + 4y² + 16x² – 12xy – 16yz + 24xz

4. Is this an identity?

(a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c²

Sol.

Question 4.
Is this an identity?
(a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c².
Solution:
No, this is not an identity.
Expanding the LHS of the equation, we get
a² + b² + c² + 2ab – 2ac – 2bc + a² + b² + c² – 2ab + 2ac – 2bc + a² + b² + c² – 2ab – 2ac + 2bc = 3a² + 3b² + 3c² – 2bc – 2ab – 2 ac which is not equal to RHS.
Hence, the given equation is not an identity.

In Grade 8, you were introduced to yet another identity, a² – b² = (a + b) (a – b)

This can be quite useful if it is rewritten as 

a² = (a + b)(a – b) + b²

Look at the following figure. Justify the identity 

a² = (a + b)(a – b) + b² for yourself.

In 750 CE, this identity was proposed by Śhrīdharāchārya as a method to quickly compute the squares of numbers. 

For example,

55² 

= (55 + 5)(55 – 5) + 5² 

= 60 × 50 + 25

= 3000 + 25 

= 3025. 

Think and Reflect

1. Try to evaluate the following using a suitable identity:

(i) 35²

(ii) 65²

(iii) 85²

(iv) 105²

Do you observe any interesting pattern?

Sol.1

Identity:

(a + b)^ 2 = a ^ 2 + 2ab + b ^ 2

(i) 35 ^ 2

35 ^ 2 = 

(30 + 5) ^ 2 

= 30 ^ 2 + 2(30)(5) + 5 ^ 2 

= 900 + 300 + 25 

=  1225

(ii) 65 ^ 2

65 ^ 2 

= (60 + 5) ^ 2 

= 3600 + 600 + 25 

= 4225

(iii) 85 ^ 2

85 ^ 2 

= (80 + 5) ^ 2 

= 6400 + 800 + 25 

= 7225

(iv) 105 ^ 2

105 ^ 2 

= (100 + 5) ^ 2 

= 10000 + 1000 + 25 

= 11025

2. Observe the two rows of figures below. They represent an algebraic identity. 

Try to identify it.

The figures represent the identity:

[{(a+b+c)²+(a+b–c)²+(a-b+c)²+(a–b–c)² =4(a² + b² + c²)}]

How to identify it

From the first row, the four squares have sides:

[ a+b+c, a+b–c, a–b+c, a–b–c ]

Therefore, their total area is

[(a+b+c)²+(a+b–c)²+(a–b+c)²+(a–b–c)²]

From the second row, the large square has side (2a), so its area is

[ (2a)² = 4a²]

Similarly, for (2b) and (2c) squares give
[ 4b², 4c²]

Hence total area is
[ 4a² + 4b² + 4c² = 4 (a² + b² + c²)]

Final identity

[(a+b+c)²+(a+b–c)²+(a–b+c)²+(a–b–c)² = 4 (a² + b² + c²)]

This is the algebraic identity represented by the given figures.

4.5 FACTORISATION USING ALGEBRA TILES

Consider a rectangle with sides x + 3 and x + 4 units.

We know that the area of such a rectangle is (x + 3)(x + 4) sq. units.

Using distributivity, we get 

(x + 3)(x + 4) 

= x² + 3x + 4x + 12 

= x² + 7x + 12

Figure helps to visualise the product of X + 3 and x + 4 using algebra tiles. The expression x + 3 is represented using an x-tile and three unit tiles. 

Similarly x + 4 is represented using an x-tile and four unit tiles. The product of these two linear factors is shown in the inner rectangle comprising the x² tile, the 7 x-tiles, and 12 unit tiles.

Note that the 7x in x² + 7x + 12 has been split as 3x + 4x. This is represented by the fact that three x-tiles are placed on the right side of the x² - tile and four x-tiles are arranged below it. Also, the 12 unit tiles are arranged in a 3 by 4 array. Once the rectangular arrangement is formed, we observe that the dimensions of the rectangle are x + 3 and x + 4 units respectively.

Think and Reflect

Suppose 7x is split as 2x + 5x can a similar rectangular arrangement be formed? Consider other possibilities and check.

Sol. 

Yes. A similar rectangular arrangement can be formed when 7x is split as 2x + 5x .

For example, consider a rectangle of length 7x and breadth Y. Split the length into 2x and 5x:

7x = 2x + 5x

Its total area can be written as

7xy = (2x + 5x)×y

= 2xy + 5xy

So the rectangular arrangement represents the distributive identity:

a(b+c) = ab + ac

Other possibilities

Since 7x can be split in several ways:

7x = x + 6x

7x = 3x + 4x

7x = 4x + 3x

7x = 5x + 2x

7x = 6x + x

All of these can form similar rectangular arrangements. The order of the two parts does not affect the total area.

Fig. 4.7 helps us to visualise two algebraic identities:

(i) The linear expressions x + 3 and x + 4 can be multiplied to obtain the identity (x + 3)(x + 4) = x² + 7x + 12

(ii) Also the expression x² + 7x + 12 can be factored into the linear factors (x + 3) and (x + 4) giving the same identity x² + 7x + 12 = (x + 3)(x + 4)

Think and Reflect

Algebra tiles can be used to represent products and find factors.

1. Figure out the product of x + 2 and x + 3 using algebra tiles.

Sol.

Product of x + 2 and x + 3

Using algebra tiles:

xx = x ^ 2

2x = 2x

3x = 3x

2 * 3 = 6

So,

(x + 2)(x + 3) = x ^ 2 + 2x + 3x + 6 (x + 2)(x + 3) = x ^ 2 + 5x + 6

2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.

Sol.

Factor x ^ 2 + 11x + 30 1x ^ 2 tile

Sum = 11

Those numbers are 5 and 6.

Using algebra tiles, arrange:

11 x-tiles

30 unit tiles

They form a rectangle with dimensions:

(x + 5) and (x + 6)

Therefore,

x ^ 2 + 11x + 30 = (x + 5)(x + 6)

Think and Reflect

We have seen that (x + 3)(x + 4) = x² + 7x + 12 Also (x + 6)(x + 7) = x² + 13x + 42 

Generalise the pattern to get an expression for (x + a)(x + b)

Sol.

We observe:

(x + 3)(x + 4) 

= x ^ 2 + 7x + 12

Here, 7 = 3 + 4 and 12 = 3 * 4

Similarly,

(x + 6)(x + 7) 

= x ^ 2 + 13x + 42

Here, 13 = 6+7 and 42 = 6 * 7

Therefore, in general:

(x + a)(x + b) 

= x ^ 2 + (a + b) * x + ab

Using distributive property:

(x + a)(x + b) 

= x(x + b) + a(x + b)

= x ^ 2 + bx + ax + ab

= x ^ 2 + (a + b) * x + ab

This is the generalised pattern.

Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + underline underline 1 , as shown in Figure What can you say about its area (2x + 3)(3x + 1)

Fill in the blanks with the appropriate expressions to make the equation true.

(px + a)(qx + b) =(___)x² +(___)x + ____

Also, verify your answer using the distributive property.

4.6 FACTORISATION WITHOUT USING ALGEBRA TILES

Consider the algebraic expression x² + 7x + 12 which was obtained by multiplying the linear terms x + 3 and x + 4 We also saw in Figure that 7x was split as 3x + 4x so that a rectangle could be formed using the algebra tiles. But how do we achieve this 'splitting of the x term' without using tiles?

Example 10: Let us begin with x² + 7x + 12 = x² + (a + b) x + ab

Comparing the coefficients of the x-term and the constant terms on both sides of the equation, we get a + b = 7 and ab = 12 Note that this is only possible when a = 3 and b = 4 or a = 4 and b = 3 

Thus, we choose one of these two possibilities and write the factors as (x + a)(x + b) that is, (x + 3)(x + 4)

Example 11: Let us try to factor x² + 11x + 30 in a similar manner. x² + 11x + 30 = x² + (a + b) x + ab

Thus, 

x² + 11x + 30 

= x² + (5 + 6) x + 30 

= x² + 5x + 6x + 30 

= x(x + 5) + 6(x + 5) 

= (x + 5)(x + 6) .

Example 12: In order to factor x² – 5x + 6 we first note that the coefficient of x is negative. Once again comparing x² – 5x + 6 with x² + (a + b) x + ab we get a + b = –5 and ab = 6 

Note that these two equations can be satisfied together only when a = – 2 and b = – 3 or vice-versa.

EXERCISE SET 4.4

1. Fill in the blanks to complete the following identities:

(i) s² – 11s + 24 = (______) (______)

(ii) (______) (x + 1) = (3x ^ 2 - 4x - 7)

(iii) 10x² – 11x – 6 =(2x – __)(___ + 2)

(iv) 6x² + 7x + 2 = (______) (______)

Sol.

(i) s² – 11s + 24 = (__________)(__________)
Solution:
s² – 11s + 24
Comparing it with s² + (a + b)s + ab, we get a + b = -11 and ab = 24.
Clearly, these two equations can be satisfied together only when a = -8 and b = -3, or vice versa.
So, s² + [(-8) + (-3 )]s + 24 = [s + (-8)][s + (-3)]
= (s – 8)(s – 3)
[x² + (a + b)x + ab = (x + a)(x + b)]
Thus, s² – 11x + 24 = (s – 8) (s – 3)

(ii) (__________) (x + 1) = (3x² – 4x – 7)

Solution:
(__________) (x + 1) = (3x² – 4x – 7)
We have, 3x² – 4x -7
Multiply the coefficient of x² (which is 3) by the constant term (which is -7): 3 x (-7) = -21
Now, we need to find two numbers that multiply to -21 and add up to the coefficient of x (which is -4).
The two numbers that satisfy this condition are -7 and 3 because (-7) x 3 = —21,
(-7) + 3 = -4
Therefore, (3x² – 4x -7) = 3x² – (7 – 3)x – 7
= 3x² + 3x – 7x – 7
= 3x(x + 1) – 7(x + 1)
= (3x – 7) (x + 1)
Thus, (3x – 7) (x + 1) = (3x² – 4x – 7)

(iii) 10x² – 11x – 6 = (2x – __) (__ + 2)
Solution:
10x² – 11x – 6 = (2x – __) (__ + 2)
We have, 10x² – 11x – 6
Multiply the coefficient of x² (which is 10) by the constant term (which is -6): 10 x (-6) = -60.
Now, we need to find two numbers that multiply to -60 and add up to the coefficient of x (which is -11).
The two numbers that satisfy this condition are -15 and 4 because (-15) x 4 = -60,
(-15)+ 4 = -11.
Therefore, 10x² – 11x – 6 = 10x² + [4 + (-15)]x – 6
= 10x² + 4x — 15x – 6
= 2x(5x + 2) – 3(5x + 2)
= (2x – 3)(5x + 2)
Thus, 10x² – 11x – 6 = (2x – 3) (5x + 2)

(iv) 6x² + 7x + 2 = (__________)(__________)
Solution:
6x² + 7x + 2 = (__________)(__________)
Multiply the coefficient of x² (which is 6) by the constant term (which is 2): 6 × 2 = 12.
Now, we need to find two numbers that multiply to 12 and add up to the coefficient of x (which is 7).
The two numbers that satisfy this condition are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.
Therefore, 6x² + 7x + 2 = 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1).
Thus, 6x² + 7x + 2 = (3x + 2)(2x + 1)

2. Select and use the identity that will help you to find the following products without multiplying directly:

(i) (41)²

(ii) (27)²

(iii) (23 x 17)

(iv) (135)²

(v) (97)²

(vi) (18 x 29)

(vii) (34 x 43)

(viii) (205)²

Sol

Question 2.
Select and use the identity that will help you to find the following products without multiplying directly:
(î) (41)²
Solution:
(41)² = (41 + 1)(41 – 1)+ 1²
=42 × 40 + 1
= 1681
[∵ a² = (a + b)(a – b) + b²]

(ii) (27)²
Solution:
(27)² = (30 – 3)² = 302² × 30 × 3 + 32
=900 – 180 + 9
=729
[∵ (a – b)² = a² – 2ab + b²]

(iii) (23 × 17)
Solution:
(23 × 17) = (20 + 3) (20 – 3) = 20² – 3²
= 400 – 9
= 391
[∵ a² – b² = (a + b)(a – b)]

(iv) (135)²
Solution:
(135)² = (140 – 5)²
= 140² – 2 × 140 × 5 + 5²
= 19600 – 1400 + 25
= 18225
[∵ (a – b)² = a² – 2ab + b²]

(v) (97)²
Solution:
(97)² = (97 + 7) (97 – 7) + 7²
= 104 × 90 + 49
= 9409
[∵ a² = (a + b)(a – b) + b²]

(vi) (18 × 29)
Solution:
(18 × 29) = (20 – 2) (20 + 9)
= 20² + (-2 + 9)20 + (-2) × 9
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= 400 + 140 – 18
= 522

(vii) (34 × 43)
Solution:
= (40 – 6) (40 + 3)
= 40² + (-6 + 3)40 + (-6) × 3
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= 40² + (-6 + 3)40 + (-6) × 3
= 1600 – 120 – 18
= 1462

(viii) (205)²
Solution:
(205)² = (205 + 5) (205 – 5) + 5²
= 210 × 200 + 25
= 42025
[∵ a² = (a + b)(a – b) + b²]

3. Factor the following:

(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc

(ii) 16s² + 25t² – 40st

(iii) r² – r – 42

(iv) 49g² + 14gh + h²

(v) 64u² + 121v² + 4w² – 176uv – 32uw + 44vw

Sol.

Question 3.
Factor the following:
(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc
Solution:
9a² + b² + 4c² – 6ab + 12ac – 4be
= (3a)² + (-b)² + (2c)² + 2(3a) (-b) + 2(3a)(2c) + 2(-b)(2c)
= (3a – b + 2c)²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

(ii) 16s² + 25t² – 40st?
Solution:
16s² + 25t² – 40st
= (4s)² + (5t)² – 2 × (4s)(5t)
= (4s – 5t)²
[∵ (a – b)² = a² + b² – 2ab]

(iii) r² – r – 42
Solution:
r² – r – 42
= r² + (-7 + 6)r + (—7)(6)
= [r+ (-7)] (r + 6)
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= (r – 7)(r + 6)

(iv) 49g² + 14gb + b²
Solution:
49g² + 14gb + b²
= (7g)² + 2(7g)b + h²
= (7g + h)² [∵ (a + b)² = a² + b² + 2ab]

(v) 64u² + 121 v² + 4W² – 176uv – 32uw + 44vw
Solution:
64u² + 121 v² + 4W² – 176uv- 32uw + 44vw
= (-8u)² + (11 v)² + (2w)² + 2(-8u)(11 v) + 2(-8u)(2v) + 2(11v) (2w)
= [(-8)u + 11v + 2w]²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= (11v + 2w – 8u)²

Think and Reflect

James and Reshma were talking about algebraic identities they learnt in school.

James: (a – b)² (a + b) = (a² – 2ab + b²)(a + b)

Reshma: I have a different idea. 

(a – b)² (a + b) = (a – b)[(a – b)(a + b)] 

                          = (a – b)(a² – b²)

I will find this product to get the answer.

According to you, who is correct and why?

Try to combine more such identities and find new results.

FINDING NEW IDENTITIES

In this section we will play around with algebraic expressions to see if we can arrive at new identities.

What do you think (a + b)³ will look like?

Let us find out the answer using the distributive property: 

(a + b)³ = (a + b)(a² + 2ab + b²) 

              = a³ + 3a³b + 3ab² + b³

Here we have found a new identity, namely, 

(a + b)³ = a³ + 3a³b + 3ab² + b³

Let us try to visualise this identity. We saw that a square of side a + b can be divided into squares and rectangles. 

This led us to the identity, 

(a + b)² = a² + 2ab + b² 

What if we have a cube of edge a +b? Can we divide a cube of edge (a + b) into smaller cubes and cuboids and represent this new identity?

We know that the volume of this cube is (a + b)³ Let us split this cube into smaller cubes and cuboids.

(a + b)³ = a³ + 3a³b + 3ab² + b³

Notice that the larger cube can be split into two cubes and six cuboids. The cubes have volumes, a³ and b³ cubic units, respectively. Amongst the other six cuboids, three have dimensions a units x a units b units and other three have dimensions a units b units x b units, and so their volumes are a²b and ab² cubic units, respectively. Hence the total volume of the six cuboids is 3a²b + 3ab².

This gives us, 

(a + b)³ = (a + b)(a² + 2ab + b²) 

              = a³ + 3a³b + 3ab² + b³

Here we have found a new identity, namely, 

(a + b)³  = a³ + 3a³b + 3ab² + b³

What happens when we replace 'b' with'–b' in this new identity? 

[a + (–b)]³ = a³ + 3a²(–b) + 3a(– b)² + (–b)³ 

This leads to the identity 

(a – b)³  = a³ – 3a³b + 3ab² – b³ 

Note that out of the four terms, two are positive and two are negative. They appear alternately in the expression.

Let us see how we can use these new identities.

Example 13: What is the side of the cube whose volume is p³ + 6p²q + 12pq² + 8q³ cubic units? 

Comparing p³ + 6p²q + 12pq² + 8q³  with the right side of the identity 

(a + b)³  = a³ + 3a³b + 3ab² + b³ 

we may rewrite it as 

(p)³ + 3(p)²(2q) + 3(p)(2q)² + (2q)³ 

which is (p + 2q)³ 

so that a = p and b = 2q. 

Hence a side of the cube will be p + 2q units.

Example 14: Now consider the expression 8n³ – 60n²m + 150nm² – 125m³. If you write it in the form (a – b)³, what will be a and b?

Sol. 

8n³ – 60n²m + 150nm² – 125m³ 

= (2n)³ – (2n)²(5m) + 3(2n)(5m)² – (5m)³  comparing it with 

a³ + 3a³b + 3ab² + b³ = (a – b)³ 

we get 

= (2n 5m). 

Hence a = 2n and b = 5m

Now let us play with known identities to discover more identities. Try to multiply the following using the distributive property.

1. (x – y)(x² + xy + y²)

2. (x + y)(x² – xy + y²)

Let us work out the product of the first one. 

(x – y)(x² + xy + y²) 

= x³ + x²y + xy² – x²y – xy² – y³ 

= x³ + x²y + xy² – x²y – xy² – y³ 

= x³ – y³

Thus (x – y)(x² + xy + y²) = x³ – y³

This is also an identity! Verify this for yourself by choosing different values of x and y. Predict what (x + y)(x² + xy + y²) will be.

Think and Reflect

We already know that 

x² – y² = (x – y)(x + y) 

Further, we have verified that 

x³ – y³ = (x + y)(x² + xy + y²) 

Observe that x – y is a common factor of 

x² – y² and x³ – y³. 

Do you think x – y is also a factor of x¹ – y¹? Note that 

x⁴ – y⁴ 

= (x²)²  – (y²)² 

= (x²  – y²)(x² + y²) 

= (x – y)(x + y)(x² + y²)

Can you see how (x – y) is a factor of x² – y² ?

How about x⁵ – y⁵ ? Does this also have x – y as a factor?

Exploring further, let us multiply (x + y + z) and (x² + y² + z² – xy – xz – yz) 

= (x + y + z) (x² + y² + z² – xy – xz – yz) 

= x (x² + y² + z² – xy – xz – yz)  + y ( x² + y² + z² – xy – xz – yz) +z (x² + y² + z² – xy – xz – yz) 

= (x³ + xy² + xz² – x²y – x²z – xyz)  + ( x²y + y³ + yz² – xy² – xyz – y²z) + (x²z + y²z + yz² – xyz – xz² – yz²) 

= x³ + xy² + xz² – x²y – x²z – xyz + x²y + y³ + yz² – xy² – xyz – y²z + x²z + z³ + yz² – xyz – xz² – yz²

= x³ + y³ + z³ + xy² – xy²  + xz² – xz² – x²y + x²y + x²z – x²z  + yz² + yz²  – y²z  – yz² – xyz – xyz – xyz

= x³ + y³ + z³ – 3xyz 

So. 

(x+y+z) (x²+y²+z²–xy–xz–yz)=x³+y³+z³–3xyz

Or

x³+y³+z³–3xyz=(x+y+z) (x²+y²+z²–xy–xz–yz) 

This is yet another identity. Let us explore some of its applications.

Example 15: The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. 

Sol.

Try to use the previous identity to find the sum of the cubes of these three numbers. Let the three numbers be x, y and z, respectively. According to the problem 

x + y + z = 10 

xyz = 25 and 

x² + y² + z² = 38 

Substituting in the identity 

(x+y+z) (x²+y²+z²–xy–xz–yz)=x³+y³+z³–3xyz

we get 

(10) (38 – xy – yz – yx) = x³+y³+z³–3(25) 

Thus, x³+y³+z³ = 380 – 10(xy + xz + yz) + 75 

x³ + y³ + z³ = 455 – 10(xy + xz + yz) ........(1)

To find (xy + xz + yz), 

we need to use the identity 

(x + y + z)² = x² + y² + z² + 2xy + 2xz + 2yz 

(x + y + z)² = x² + y² + z² + 2(xy + xz + yz)

We get 

100 = 38 + 2(xy + xz + yz) 

100 –38 = 2(xy + xz + yz) 

62 = 2(xy + xz + yz) 

(xy + xz + yz) = 62/2

xy + xz + yz = 31

Thus xy + xz + yz = 31 

Now, substituting this into our earlier equation (1), we obtain 

x³ + y³ + z³ = 455 – 10(xy + xz + yz)

                    = 455 – 10 (31) 

                    = 455 – 310 

x³ + y³ + z³ = 145

4.8 SIMPLIFYING RATIONAL EXPRESSIONS

In the earlier sections of this chapter, we have learnt how to factorise algebraic expressions. Let us see how we can simplify some rational algebraic expressions using factorisation.

Example 16: Simplify the rational expression assuming that 5x² + 5x –100 ≠ 0. 

     (x² – 7x + 12)

=  ________________

     (5x² + 5x – 100)

Sol.

     (x² – 7x + 12)

=  ________________

     (5x² + 5x – 100)

     (x² – 4x  –3x + 12)

=  ________________

     5(x² + x – 20)

       (x² – 4x  –3x + 12)

=  ______________________

      5 (x² + 5x –4x – 20)

       x(x – 4) – 3(x – 4)

=  ______________________

      5 [ x(x + 5) –4 (x + 5)]

       (x – 4) (x – 3)

=  ______________________

      5 [ (x + 5) (x – 4)]

       (x – 4) (x – 3)

=  ________________

      5 (x + 5) (x – 4)

       (x – 3)

=  __________

      5 (x + 5)

Think and Reflect

Try to simplify the following rational expression:

     (36s² - 12st + t²)         (6s – t)²

   ___________________ =  ______________

       (t² + 2ts - 48s²)          (______ )(______ )

(Hint: Factor and simplify the rational expressions t² + 2ts - 48s² assuming that t² + 2ts – 48s² ≠ 0

EXERCISE SET 4.5

1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i) (3p² – 3pq –18q²)/(p² + 3pq – 10q²)

(ii) (n³–3n²m + 3nm³ – m³)/(5m²–10mn + 5n²)

(iii) w³–v³ + x³ +3wvx w² + v²+x²-2wv-2vx+2wx

(iv) (4y² – 20yz + 25z²)/(25z² – 4y²)

(v) ((x² + x – 6)(x² – 7x + 12))/((x² – 6x + 8)(x² – 9))

(vi) (p⁴ – 16)/(p² – 4p + 4)

Sol.

(i) 3p2−3pq−18q2p2+3pq−10q2
Solution:
3p2−3pq−18q2p2+3pq−10q2
We have,
3p² – 3pq – 18q² = 3 (p² – pq – 6q²)
= 3(p² – 3pq + 2pq – 6q²)
(∵ -6 = -3 × 2 and -3 + 2 = -1)
= 3[p(p – 3q) + 2q(p – 3q)]
3p² – 3pq – 18q² = 3(p + 2q)(p – 3q) …(i)
Now, p² + 3pq – 10q² = p² + 5pq – 2pq – 10q²
[∵ -10 = 5 × (-2) and 5 + (-2) = 3]
= p(p + 5q) – 2q(p + 5q)
Therefore, p² + 3pq – 10q² = (p – 2q)(p + 5q) …(ii)
From (i) and (ii), we get
3p2−3pq−18q2p2+3pq−10q2=3(p+2q)(p−3q)(p−2q)(p+5q)

(ii) n3−3n2m+3nm2−m35m2−10mn+5n2
Solution:
Given the expression: n3−3n2m+3nm2−m35m2−10mn+5n2
Step 1: Factorise the numerator and denominator.
Numerator: The expression n³ – 3n²m + 3nm² – m³ is a perfect cube expansion and factors as (n – mf
Denominator: The expression 5m² – 10mn + 5n² factors as 5(m – n)²
Step 2: Simplify the expression.
Now, the expression becomes (n−m)35(m−n)2
Since (n – m) = -(m – n), we get
−(m−n)35(m−n)2=−(m−n)5=n−m5

(iii) w³−v³+x³+3wvx /w²+v²+x²−2wv−2vx+2wx
Solution:
We have,
w³−v³+x³+3wvx /w²+v²+x²−2wv−2vx+2wx
Using the identity 

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² -ab – bc – ca),
take a = w ,b = -v, c = x, then
w³ – v³ + x³ + 3wvx = (w – v + x) (w² + v² + x² + wv + vx – wx)

Also, the denominator
w² + v² + x² – 2wv – 2vx + 2wx 

= (w – v + x)²
Hence,
= (w−v+x)(w²+v²+x²+wv+vx−wx)

= (w−v+x)²
= (w²+v²+x2+wv+vx−wx)/ (w−v+x)

(iv) 4y²−20yz+25z² /(25z²−4y²)
Solution:
4y²−20yz+25z²/(25z²−4y²)

For Numerator 

4y² – 20yz + 25z² 

= (2y)² – 2 × 2y × 5z + (5z)² 

= (2y – 5z)²

For Dinominator

25z² – 4y² 

= (5z)² – (2y)²

= (5z – 2y)(5z + 2y) 

Now

4y²−20yz+25z² / (25z²−4y²)

= (2y−5z)² / (5z−2y)(5z+2y)
Since (2y – 5z) = -(5z – 2y), (2y – 5z)² = (5z – 2y)²
So, (5z−2y)2(5z−2y)(5z+2y)=5z−2y5z+2y

(v) (x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)
Solution:
We have,
(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)
Now, numerator = (x² + x – 6)(x² -7x + 12)
= [x² + 3x – 2x – 6].[x² – 4x – 3x + 12]
= [x(x + 3) – 2(x + 3)].[x(x – 4) -3(x – 4)]
= [(x + 3)(x – 2)].[(x – 4)(x – 3)]
and denominator = (x² – 6x + 8)(x² – 9)
= [x² – 4x – 2x + 8].(x – 3)(x + 3)
[va² – b² = (a + b)(a – b)]
= [x(x – 4) – 2(x – 4)] .(x – 3)(x + 3)
= (x – 4) (x – 2) (x – 3) (x + 3)
Now,
(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)
= (x+3)(x−2)(x−3)(x−4)(x−2)(x−4)(x−3)(x+3) = 1

(vi) p4−16p2−4p+4
Solution:

Example 17: Saira has arranged a square of side x units, 8 rectangular strips of sides x units and width 1 unit, and 15 squares of side 1 unit to form a bigger rectangle. Find the length and breadth of the rectangle in terms of x.

Area covered by a square of side x units = x² sq. units

Area covered by rectangle of sides x units and 1 unit = x sq. units

Total area covered by 8 such rectangles = 8x sq. units

Area covered by a square of side 1 unit 1 sq. units

Total area covered by 15 such squares = 15 sq. units

Total area covered by all the squares and rectangles

= x² + 8x + 15 sq. units.

Area of Saira's rectangle = x² + 8x + 15 sq. units.

We can factor x² + 8x = 15 to obtain the dimensions of the rectangle prepared by Saira.

For factoring x² + 8x + 15 we need a and b such that a + b = 8 and ab = 15 

So, a = 3 and b = 5 is one possibility. 

The possible length and breadth of Saira's rectangle in terms of x are: length = x + 5 units and breadth = x + 3 units.

Now draw Saira's rectangle using these pieces.

Example 18: A rectangular pool is such that its breadth is 4 metres less than its length and its area is 96 sq. metres. Find the length and breadth of the pool.

Let the length of the pool be x metres. Then the breadth of the pool is x – 4 metres.

88

For factoring x ^ 2 + 8x + 15 we need a and b such that a + b = 8 and ab = 15

So, a = 3 and b = 5 is one possibility. The possible length and breadth of Saira's rectangle in terms of x are: length = x + 5 units and breadth = x + 3 units.

Now draw Saira's rectangle using these pieces.

Example 18: A rectangular pool is such that its breadth is 4 metres less than its length and its area is 96 sq. metres. Find the length and breadth of the pool.

Let the length of the pool be x metres. Then the breadth of the pool is x - 4 metres.

Since the area is given to be 96 sq. metres, we get x(x - 4) = 96

x ^ 2 - 4x = 96 or x ^ 2 - 4x - 96 = 0

We choose appropriate factors of -96 to split the term -4x. We know that, (- 12) * 8 = 96 and (-12)+8-4. Hence - 4x can be written as - 12x + 8x x ^ 2 - 4x - 96 = x ^ 2 - 12x + 8x - 96 = x(x - 12) + 8(x - 12) = (x - 12)(x + 8) = 0 .

Thus, x² – 4x – 96 = 0 implies that (x – 12)(x + 8) = 0

This means either x –12 = 0 or x + 8 = 0

That is, x = 12 or x = – 8 Since x is the length of the pool, it cannot be negative. Therefore, we ignore x = – 8 and x = 12 metres must be the length of the pool. The breadth of the pool = x – 4 = 12 – 4 = 8 metres.

END-OF-CHAPTER EXERCISES

1. Use suitable identities to find the following products:

(i) (–3x + 4)²

(ii) (2s + 7)(2s – 7)

(iii) (p² + 1/2)(p² – 1/2) 

(iv) (2n + 7)(2n – 7)

(v) (s – 2t)(s² + 2st + 4t²)

(vi) (1/(2r) – 4r)²

(vii) (–3m + 4k –l)²

(viii) (x – 1/3 y)³

(ix) (7/2 k – 2/3 m)³

Sol.

Question 1.
Use suitable identities to find the following products:
(i) (-3x + 4)²
Solution:
(-3x + 4)²
We can expand this using the identity
(a + b)² = a² + 2ab + b²
(-3x + 4)² = (-3x)² + 2(-3x)(4) + 4²
= 9x² – 24x + 16

(ii) (2s + 7) (2s – 7)
Solution:
(2s + 7) (2s – 7)
We can use the identity
(a + b)(a – b) = a² – b²
(25 + 7) (25 – 7) = (25)² – (7)²
= 4s² – 49

(iii) (p² + 12)(p² – 12)
Solution:
(p² + 12)(p² – 12)
Using the identity: (a + b)(a – b) = a² – b²
(p² + 12)(p² – 12)
= (p²)² – (12)²
= p⁴ – 14

(iv) (2n + 7) (2n – 7)
Solution:
(2n + 1)(2n – 1)
Using the identity: (a + b)(a – b) = a² – b²
(2n + 7) (2n – 7) = (2n)² – 7²
= 4n² – 49

(v) (s – 2t) (s² + 2st + 4t²)
Solution:
(s – 2t) (s² + 2st + 4t²)
Here, we use the identity
(x – y)(x² + xy + y²) = x³ – y³
(s – 2t) (s² + 2st + 4t²) = s² – (2t)²
= s³ – 8t³

(vi) (12r – 4r)²
Solution:
(12r – 4r)²
We simplify this by using the identity
(a – b)² = a² – 2ab + b²
(12r)² – 2(12r)(4r) + (4r)²
= 14r2 – 4 + 16r²

(vii) (-3m + 4k – l)²
Solution:
(-3m + 4k – l)²
We can expand using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
(-3m + 4k – l)² = (—3m)² + (4k)² + (-l)² + 2(-3m)(4k) + 2(-3m)(-l) + 2(4k)(-l)
= 9m² + 16k² +12 — 24mk + 6ml – 8kl

(viii) (x – 13)³
Solution:
(x – 13)³
To expand this, we use the identity:
(a – b)³ = a³ – 3a²b + 3ab² – b³
= a³ – b³ – 3ab(a – b)
(x – 13)³ = x³ – 127y³ – 3x x 13y(x – 13y)
= x³ – 127y – x²y + x3

(ix) (72k – 23m)³
Solution:
To expand this, we use the identity:
(a – b)³ = a³ – 3a²b + 3ab² – b³

2. Find the values using suitable identities:

(i) 17 × 21

(ii) 104 × 96

(iii) 24 × 16

(iv) 147³

(vi) 127³

(v) 199³

(viii) (– 299)³

(vii) (– 107)³

Sol.

Question 2.
Find the values using suitable identities:
(i) 17 × 21
Solution:
17 × 21
Here, 17 = 20 – 3 and 21 = 20 + 1, so:
17 × 21 = (20 – 3)(20 + 1)
Using the identity (x + a)(x + b)
= x² + (a + h)x + ab
= 20² + (-3 + 1)20 + (-3)(1)
= 400 – 40 – 3
= 357

(ii) 104 × 96
Solution:
104 × 96
Here, 104 = 100 + 4 and 96 = 100 – 4,
So, 104 × 96 = (100 + 4)(100 – 4)
Using the identity (a + b)(a – b) = a² – b²
= 100² – 4²
= 10000 – 16
= 9984

(iii) 24 × 16
Solution:
24 × 16
Here, 24 = 20 + 4 and 16 = 20 – 4
So, 24 × 16 = (20 + 4)(20 – 4)
Using the identity (a + b)(a – b) = a² – b²
= 20² – 4²
= 400- 16
= 384

(iv) 147³
Solution:
147³
We can use the identity for cubes:
(a – b)³ = a³ – 3a2b + 3 ab2 – b³
Here, 147 = 150 – 3,
So, 147³ = (150 – 3)³
= 150³ – 3 × 150² × 3 + 3 × 150 × 3² – 3³
Simplifying:
= 3375000 – 202500 + 4050 – 27
= 3176523

(v) 199³
Solution:
199³
We can use the identity for cubes:
(a – b)³= a³ – 3a²b + 3ab² – b³
Here, 199 = 200- 1,
So, 199³ = (200 – 1)³
= 2003 – 3 × 200² × 1 + 3 × 200 × 1² – 1³
= 8000000 – 120000 + 600 – 1
= 7880599

(vi) 127³
Solution:
127³
We can use the identity for cubes:
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, 127 = 130 – 3,
So, 127³ = (130 – 3)³
= 130³ – 3 × 130² × 3 + 3 × 130 × 3² – 3³
= 2197000 – 152100 + 3510 – 27
= 2048383

(vii) (-107)³
Solution:
(–107)³

(–100 – 7)³
We can use the identity for cubes:
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, –107 =–100 – 7,
So, (–100 – 7)³
= (–100)³ – 3 × (–100)² × 7 + 3 × (–100) × 7² – 7³
= –1000000 – 210000 + (–14700) – 343
= –1225043

(viii) (–299)³
Solution:
(–299)³

(–300 + 1)³
We can use the identity for cubes:
(a + b)³ = a³ + 3a²b + 3 ab² + b³
So, (–300 + 1)³
= (–300)³ + 3 × (–300)² × 1 + 3 × (–300) × 1²+ 1³
= –27000000 + 270000 – 900 + 1
= – 26730899

3. Factor the following algebraic expressions:

(i) 4y² + 1 + 1/16y²

(ii) 9m² – 1/(25n²)

(iii) 27b³ – 1/64b³

(iv) x² + 5x/6+16

Sol.

(i) 4y² + 1 + 1/16y²
Solution:
4y² + 1 + 1/16y2
= (2y)² + 2 × (2y) × (14y)+(14y)²
= (2y + 12y)²
[∵ (a + b)² = a² + 2ab + b²]

(ii) 9m² – 1/25n2
Solution:
9m² – 1/25n2
= (3m)² – (1/5n)²
= (3m + 1/5n)(3m – 1/5n)
[∵ a² – b² = (a + b)(a – b)]

(iii) 27b³ – 1/64b³
Solution:

(iv) x² + 5x/6+16
Solution:

(v) 27u³ – 1/125−27u²/5+9u25
Solution:

(vi) 64y³ – 1125z³
Solution:

(vii) p³ + 27q³ + r³ – 9pqr
Solution:
p³ + 27q³ + r³ – 9pqr
We have, x³ + y³ + z³ – 3xyz
= (x + y + z)(x³ + y³ + z³ – xy – xz – yz)
= (p + 3q + r)(p³ + 9q³ + r³ – 3pq – 3qr – pr)

(viii) 9m² – 12m + 4
Solution:
9m² – 12m + 4
= (3m)² – 2(3m)(2) + (2)²
= (3m – 2)²
[∵ (a – b)² = a² – 2ab + b²]

(ix) 9x³ – 83 y³ + z33 + 6xyz
Solution:
9x³ – 8/3 y³ + z/3³ + 6xyz
= 1/3(27x³ – 8y³ + z³ + 18xyz)
= 13[(3x)³ + (-2y)³ + (z)³ – 3(3x)(-2y)(z)]
= 13(3x – 2y +z)(9x³ + 4y³ + z³ + 6xy + 2yz – 3xz)
[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
Solution:
4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 1(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)

(xi) 27u³ – 1/216 − 9u²/2 + u/4
Solution:
27u³ – 1/216 − 9u²/2 + u/4
= (3u)³ – (16)³ – 3(3u)²(16) + 3(3u)(16)²
= (3u – 16)³
[∵ (a – b)³ = a³ – b³ – 3a²b + 3ab²]

Question 4.
Simplify the following:
(i) 4x²+4x+1/4x2−1
(ii) 9(3a³−24b³) / 9a²−36b²
(iii) s³+125t³ / s²−2st−35t²
Note: Assume that the denominators are not equal to 0.
Solution:

Sol.

5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i) 25a² – 30ab + 9b²

(ii) 36s² – 49t²

Sol.

Question 5.
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a² – 30ab + 9b²
Solution:
25a² – 30ab + 9b²
= (5a)² – 2(5a)(3b) + (3b)²
= (5a – 3b)²
[∵ (a – b) = a² – 2ab + b²]
= (5a – 3b)(5a – 3b)
Area of a rectangle = length × breadth
Possible dimensions:
Length = (5a – 3b) units
Breadth = (5a – 3b) units

(ii) 36s² – 49t²
Solution:
36s² – 49t²
= (6s)² – (7t)²
= (6s + 7t)(6s – 7t)
[∵ a² – b² = (a + b) (a – b)]
Area of a rectangle = length × breadth
Possible dimensions:
Length = (6s + 7t) units
Breadth = (6s – 7t) units

6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) 6a² – 24b²

(ii) 3ps² – 15ps + 12p

Sol.

(i) 6a² – 24b²
Solution:
6a² – 24b²
= 6(a² – 4b²)
= 6[a² -(2b)²]
= 6(a + 2 b)(a – 2b) [∵ a² – b² = (a + b)(a – b)]
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 6 units
Breadth = (a + 2b) units
Height = (a – 2b) units

(ii) 3ps² – 15ps +12p
Solution:
3ps² – 15ps + 12p
= 3p(s² – 5s + 4)
= 3p(s² – 4s – s + 4)
= 3p[s(s – 4) – 1(s – 4)]
= 3p(s – 1)(s – 4)
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 3p units
Breadth = (s – 1) units

Height = (s – 4) units.

7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.

Sol.

Area of the playground = 40 m × 40 m
= (40)² m²
Area of the playground along with the path
= (40 + 2s)m × (40 + 2s)m
= (40 + 2s)² m²
Area of the path = (40 + 2s)² – (40)²
= (40 + 2s + 40)(40 + 2s-40) [v a² – b² = (a + b)(a – b)]
= (80 + 2s)(2s)
= 160s + 4s²
So, the area of the path is (4s² + 160s) m².

8. If a number plus its reciprocal equals 10/3 find the number.

Sol.

Let the number be x
According to the statement, we have
x + 1x=103
On multiplying both sides by lx, we get
⇒ 3x² + 3 = 10x
⇒ 3x² – 10x + 3 = 0
Now, splitting the middle term, we get
⇒ 3x² – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1) (x – 3) = 0
⇒ 3x – 1 = 0 or x – 3 = 0
⇒ x = 13 or x = 3
Hence, the required numbers are – and 3.

9. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.

Sol.

Area of the pool = 2x² + 7x + 3 square hastas
= 2x² + 6x + x + 3 square hastas
= 2x(x + 3) + 10c+ 3) square hastas
= (2x + 1)(x + 3) square hastas
Area of the pool = length × breadth
Width of the pool = (2x + 1) hastas (given)
Therefore, length of the pool = (x + 3) hastas

*10. If both x – 2 and x – 1/2 are factors of px² + 5x + r show that p = r

Sol.

px² + 5x + r
Factors = x — 2, x – 12
⇒ x = 2 or 12
For x = 2,
p(2)² + 5(2) + r = 0
4p + 10 + r = 0 …(1)

For x = 12
p(12)² + 5(12) + r = 0
p4+52 + r = 0 …….(2)

Equating (1) and (2), we get

p = -2
Substituting p = -2 in equation (1), we get
4(-2) + 10 + r = 0
⇒ -8+ 10 + r = 0
⇒ 2 + r = 0
⇒ r = -2
∴ p = r = – 2
Hence, p = r

*11. If a + b + c = 5 and ab + bc + ca = 10 then prove that a³+b³+c³–3abc = –25.

Sol

We know that
a³ + b³ + c³ – 3abc
= (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c) [a² + b² + c² – (ab + be + ca)]
= 5(a² + b² + c² – (ab + bc + ca))
= 5(a² + b² + c² – 10) …(1)
Now, a + b + c = 5
Squaring both sides, we get
(a + b + c)² = 5²
⇒ a² + b² + c² + 2(ab + bc + ca) = 25
⇒ a² + b² + c² + 2(10) = 25
⇒ a² + b² + c² = 25 – 20 = 5
Substituting a² + b² + c² = 5 in equation (1)
5(5 – 10) = 5(-5) = — 25
Hence proved.

*12. By factoring the expression, check that n³ – n is always divisible by 6 for all natural numbers n. Give reasons.

Sol

Given, n³ – n

Now, factorising the expression, we get n³ – n = n(n² – 1)

Also, using the identity a² – b² = (a – b) (a + b), we get
n(n² – 1) = n(n – 1) (n + 1)
Now, n – 1, n, n + 1 are three consecutive natural numbers, and among any three consecutive numbers, one is divisible by 2, and also one is divisible by 3.
Hence, the product (n – 1) (n + 1) is divisible by 2 × 3 = 6.
Therefore, n³ – n is always divisible by 6 for all natural numbers n.

*13. Find the value of

(i) x³ + y³ – 12xy + 64 when x + y = – 4

(ii) x³ – 8y³ – 36xy – 216 when x = 2y + 6

Sol.

x³ + y³ – 12xy – 64
Using the identity: (a + b)³ = a³ + b³ + 3ab(a + b)
(x + y)³ = x³ + y³ + 3xy(x + y)
⇒ (-4)³ = x³ + y³ + 3xy(—4)
[∵ x +y = —4]
⇒ -64 = x³ + y³ – 12xy
⇒ x³ + y³ — 12xy + 64 = 0

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6
Solution:
We are given, x = 2y + 6
⇒ x – 2y = 6
Now, x³ – 8y³ – 36xy – 216
Using the identity: (a – b)³ = a³ – b³ – 3ab(a – b)
(x – 2y)³ = x³ – (2y)³ – 3x(2y)(x – 2y)
(6)³ = x³ – 8y³ – 6xy(6)
216 = x³ – 8y³ – 36xy
⇒ x³ – 8y³ – 36xy – 216 = 0

CHAPTER SUMMARY

★ Identities are equations that are true for all values of the variables.

★ One of the ways to visualise identities is using geometrical models or algebra tiles.

★ Identities can also be used to factor algebraic expressions.

★ Factorisation of quadratic expressions may be visualised by means of algebra tiles.

★ Identities can also be used to simplify calculations such as squaring numbers or evaluating products of numbers.

★ Rational algebraic expressions may be simplified by factorisation and removing the common factors in the numerator and denominator, provided such a factor exists and it is not equal to zero.

★ We have studied the following identities in this chapter:

★ (x + y)² = x² + 2xy + y²

★ (x – y)² = x² – 2xy + y²

★ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

★ (x + y)(x – y) = x² – y²

★ (x + a)(x + b) = x² + (a + b)x + ab

★ (ax + b)(cx + d) = ac.x² + (ad + bc)x + bd

★ x³ – y³ = (x – y)(x² + xy + y²)

Ex 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.3 Solutions

Exercise 4.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.3 Solutions

Question 1.
Find the following squares using one of the above calculations easier.
(i) 117²
Solution:
117² = (100 + 10 + 7)²
= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 10000 +100 + 49 + 2000 + 140 + 1400 = 13689

(ii) 78²
Solution:
78² = (80 – 2)² = 80² – 2 × 80 × 2 + 2²
[∵(a – b)² = a² – 2ab + b²]
= 6400 – 320 + 4
= 6084

(iii) 198²
Solution:
198² = (200 – 2)²
= 200² + 2² – 2 × 200 × 2
[∵(a – b)² = a² – 2ab + b²]
= 40000 – 800 + 4
= 39204

(iv) 214²
Solution:
214² = (200 + 10 + 4)²
= 200² + 10² + 4² + 2(200)(10) + 2(10)(4) + 2(4)(200)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 40000 + 100 + 16 + 4000 + 80 + 1600
= 45796

(v) 1104²
Solution:
1104² = (1000+ 100 + 4)²
= 1000² + 100² + 4² + 2(1000) (100) + 2(100)(4) + 2(4)(1000)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2be + 2ca]
= 1000000 + 10000 + 16 + 200000 + 800 + 8000
= 1218816

(vi) 1120²
Solution:
1120² = (1000 + 100 + 20)²
= 1000² + 100² + 20² + 2(1000)(100) + 2(100)(20) + 2(20)(1000)
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= 1000000 + 10000 + 400 + 200000 + 4000 + 40000
= 1254400

Question 2.
Factor using suitable identities:
(i) 16y² – 24y + 9
Solution:
16y² – 24y + 9 = (4y)² – 2 × (4y) × 3 + (3)²
= (4y – 3)²
[∵(a – b)² = a² – 2ab + b²]

(ii) 94s² + 6st + 4t
Solution:
94s² + 6st + 4t
= (32s)² + 2 × (32s) × (2t) + (2t)²
= (32s + 2t)²

(iii) m29+mk3+k24 + 3nk + 2mn + 9n²
Solution:
= (m3)2+(k2)2 + (3n)² + 2 × (m3) × (3n) + 2 × (k2)×(m3) + 2 × (k2) × (3n)
= (m3+k2+3n)2
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

(iv) p216 – 216p2
Solution:
(p4)2 – 2 × p4×4p+(4p)2
= (p4−4p)2
[∵(a – b)² = a² – 2ab + b²]

(v) 9a² + 4b² + c² – 12ab + 6ac – 46c
Solution:
9a² + 4b² + c² – 12ab + 6ac – 46c
= (3a)² + (-2b)² + (c)² + 2(3a)(-2b) + 2(3a)(c) + 2(-2b)(c)
= [3a + (-2b) + c]²
= (3a – 2b + c)²

Question 3.
Expand the following using the identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
(i) (p + 3q + 7r)²
Solution:
(p + 3q + 7r)²
= p² + (3q)² + (7r)² + 2{p)(3q) + 2(3q)(7r) + 2(p)(7r)
= p² + 9q² + 49r² + 6pq + 42 qr + 14pr

(ii) (3x – 2y + 4z)²
Solution:
(3x – 2y + 4z)²
= (3x)² + (-2y)² + (4z)² + 2(3x)(-2y) + 2(-2y) + 2(-2y)(4z) + 2(3x)(4z)
= 9x² + 4y² + 16x² – 12xy – 16yz + 24xz

Question 4.
Is this an identity?
(a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c².
Solution:
No, this is not an identity.
Expanding the LHS of the equation, we get
a² + b² + c² + 2ab – 2ac – 2bc + a² + b² + c² – 2ab + 2ac – 2bc + a² + b² + c² – 2ab – 2ac + 2bc = 3a² + 3b² + 3c² – 2bc – 2ab – 2 ac which is not equal to RHS.
Hence, the given equation is not an identity.

Ex 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.4 Solutions

Exercise 4.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.4 Solutions

Question 1.
Fill in the blanks to complete the following identities:
(i) s² – 11s + 24 = (__________)(__________)
Solution:
s² – 11s + 24
Comparing it with s² + (a + b)s + ab, we get a + b = -11 and ab = 24.
Clearly, these two equations can be satisfied together only when a = -8 and b = -3, or vice versa.
So, s² + [(-8) + (-3 )]s + 24 = [s + (-8)][s + (-3)]
= (s – 8)(s – 3)
[x² + (a + b)x + ab = (x + a)(x + b)]
Thus, s² – 11x + 24 = (s – 8) (s – 3)

(ii) (__________) (x + 1) = (3x² – 4x – 7)
Solution:
(__________) (x + 1) = (3x² – 4x – 7)
We have, 3x² – 4x -7
Multiply the coefficient of x² (which is 3) by the constant term (which is -7): 3 x (-7) = -21
Now, we need to find two numbers that multiply to -21 and add up to the coefficient of x (which is -4).
The two numbers that satisfy this condition are -7 and 3 because (-7) x 3 = —21,
(-7) + 3 = -4
Therefore, (3x² – 4x -7) = 3x² – (7 – 3)x – 7
= 3x² + 3x – 7x – 7
= 3x(x + 1) – 7(x + 1)
= (3x – 7) (x + 1)
Thus, (3x – 7) (x + 1) = (3x² – 4x – 7)

(iii) 10x² – 11x – 6 = (2x – __) (__ + 2)
Solution:
10x² – 11x – 6 = (2x – __) (__ + 2)
We have, 10x² – 11x – 6
Multiply the coefficient of x² (which is 10) by the constant term (which is -6): 10 x (-6) = -60.
Now, we need to find two numbers that multiply to -60 and add up to the coefficient of x (which is -11).
The two numbers that satisfy this condition are -15 and 4 because (-15) x 4 = -60,
(-15)+ 4 = -11.
Therefore, 10x² – 11x – 6 = 10x² + [4 + (-15)]x – 6
= 10x² + 4x — 15x – 6
= 2x(5x + 2) – 3(5x + 2)
= (2x – 3)(5x + 2)
Thus, 10x² – 11x – 6 = (2x – 3) (5x + 2)

(iv) 6x² + 7x + 2 = (__________)(__________)
Solution:
6x² + 7x + 2 = (__________)(__________)
Multiply the coefficient of x² (which is 6) by the constant term (which is 2): 6 × 2 = 12.
Now, we need to find two numbers that multiply to 12 and add up to the coefficient of x (which is 7).
The two numbers that satisfy this condition are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.
Therefore, 6x² + 7x + 2 = 6x² + 3x + 4x + 2
= 3x(2x + 1) + 2(2x + 1)
= (3x + 2)(2x + 1).
Thus, 6x² + 7x + 2 = (3x + 2)(2x + 1)

Question 2.
Select and use the identity that will help you to find the following products without multiplying directly:
(î) (41)²
Solution:
(41)² = (41 + 1)(41 – 1)+ 1²
=42 × 40 + 1
= 1681
[∵ a² = (a + b)(a – b) + b²]

(ii) (27)²
Solution:
(27)² = (30 – 3)² = 302² × 30 × 3 + 32
=900 – 180 + 9
=729
[∵ (a – b)² = a² – 2ab + b²]

(iii) (23 × 17)
Solution:
(23 × 17) = (20 + 3) (20 – 3) = 20² – 3²
= 400 – 9
= 391
[∵ a² – b² = (a + b)(a – b)]

(iv) (135)²
Solution:
(135)² = (140 – 5)²
= 140² – 2 × 140 × 5 + 5²
= 19600 – 1400 + 25
= 18225
[∵ (a – b)² = a² – 2ab + b²]

(v) (97)²
Solution:
(97)² = (97 + 7) (97 – 7) + 7²
= 104 × 90 + 49
= 9409
[∵ a² = (a + b)(a – b) + b²]

(vi) (18 × 29)
Solution:
(18 × 29) = (20 – 2) (20 + 9)
= 20² + (-2 + 9)20 + (-2) × 9
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= 400 + 140 – 18
= 522

(vii) (34 × 43)
Solution:
= (40 – 6) (40 + 3)
= 40² + (-6 + 3)40 + (-6) × 3
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= 40² + (-6 + 3)40 + (-6) × 3
= 1600 – 120 – 18
= 1462

(viii) (205)²
Solution:
(205)² = (205 + 5) (205 – 5) + 5²
= 210 × 200 + 25
= 42025
[∵ a² = (a + b)(a – b) + b²]

Question 3.
Factor the following:
(i) 9a² + b² + 4c² – 6ab + 12ac – 4bc
Solution:
9a² + b² + 4c² – 6ab + 12ac – 4be
= (3a)² + (-b)² + (2c)² + 2(3a) (-b) + 2(3a)(2c) + 2(-b)(2c)
= (3a – b + 2c)²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]

(ii) 16s² + 25t² – 40st?
Solution:
16s² + 25t² – 40st
= (4s)² + (5t)² – 2 × (4s)(5t)
= (4s – 5t)²
[∵ (a – b)² = a² + b² – 2ab]

(iii) r² – r – 42
Solution:
r² – r – 42
= r² + (-7 + 6)r + (—7)(6)
= [r+ (-7)] (r + 6)
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= (r – 7)(r + 6)

(iv) 49g² + 14gb + b²
Solution:
49g² + 14gb + b²
= (7g)² + 2(7g)b + h²
= (7g + h)² [∵ (a + b)² = a² + b² + 2ab]

(v) 64u² + 121 v² + 4W² – 176uv – 32uw + 44vw
Solution:
64u² + 121 v² + 4W² – 176uv- 32uw + 44vw
= (-8u)² + (11 v)² + (2w)² + 2(-8u)(11 v) + 2(-8u)(2v) + 2(11v) (2w)
= [(-8)u + 11v + 2w]²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca]
= (11v + 2w – 8u)²

Ex 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 4.5 Solutions

Exercise 4.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 4.5 Solutions

Q. 1. Simplify the following rational expressions, assuming that the expressions in the denominators are not equal to zero:
(i) (3p²−3pq−18q²)/(p²+3pq−10q²)
Solution:
(3p²−3pq−18q²)/(p²+3pq−10q²)
We have,
3p² – 3pq – 18q² 

= 3 (p² – pq – 6q²)
= 3(p² – 3pq + 2pq – 6q²)
(∵ -6 = -3 × 2 and -3 + 2 = -1)
= 3[p(p – 3q) + 2q(p – 3q)]
= 3(p + 2q)(p – 3q) …(i)
Now, 

p² + 3pq – 10q² 

= p² + 5pq – 2pq – 10q²
[∵ -10 = 5 × (-2) and 5 + (-2) = 3]
= p(p + 5q) – 2q(p + 5q)
= (p – 2q)(p + 5q) …(ii)
From (i) and (ii), we get
(3p²−3pq−18q²)/(p²+3pq−10q²)

=3(p+2q)(p−3q)/(p−2q)(p+5q)

(ii) (n³−3n²m+3nm³−m³)/(5m² −10mn+5n²)
Solution:
Given the expression: n3−3n2m+3nm2−m35m2−10mn+5n2
Step 1: Factorise the numerator and denominator.
Numerator: 

The expression 

n³ – 3n²m + 3nm² – m³ is a perfect cube expansion and factors as (n – m)³
Denominator: 

The expression 

5m² – 10mn + 5n² factors as 

5(m – n)²

Step 2: Simplify the expression.
Now, the expression becomes (n−m)³ /5(m−n)²
Since (n – m) = –(m – n), we get

−(m−n)³/5(m−n)² 

=−(m−n)5

=(n−m)5

(iii) (w³−v³+x3+3wvx) /(w²+v²+x²−2wv−2vx+2wx)
Solution:
We have,
(w³−v³+x³+3wvx)/(w²+v²+x²−2wv−2vx+2wx)
Using the identity 

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² -ab – bc – ca),
take a = w, b = -v, c = x, then
w³ – v³ + x³ + 3wvx 

= (w – v + x) (w² + v² + x² + wv + vx – wx)
Also, 

The Denominator
w² + v² + x² – 2wv – 2vx + 2wx 

= (w – v + x)²
Hence,
= (w−v+x)(w²+v²+x²+wv+vx−wx)/(w−v+x)2l²
= (w²+v²+x²+wv+vx−wx)/(w−v+x)

(iv) (4y²−20yz+25z²)/(25z²−4y²)
Solution:
(4y²−20yz+25z²)/(25z²−4y²)
The Numerator 

4y² – 20yz + 25z²

= (2y – 5z)^{2 .....(1)}

The Dinominator 
25z² – 4y² 

= (5z – 2y)(5z + 2y) .....(2)

Equation (1) and (2)

So, (5z−2y)²/(5z−2y)(5z+2y)

=(5z−2y)/(5z+2y)

(v) (x²+x−6)(x²−7x+12)/(x²−6x+8)(x²−9)
Sol.
We have,
(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)
Now, numerator = (x² + x – 6)(x² -7x + 12)
= [x² + 3x – 2x – 6].[x² – 4x – 3x + 12]
= [x(x + 3) – 2(x + 3)].[x(x – 4) -3(x – 4)]
= [(x + 3)(x – 2)].[(x – 4)(x – 3)]
and denominator = (x² – 6x + 8)(x² – 9)
= [x² – 4x – 2x + 8].(x – 3)(x + 3)
[va² – b² = (a + b)(a – b)]
= [x(x – 4) – 2(x – 4)] .(x – 3)(x + 3)
= (x – 4) (x – 2) (x – 3) (x + 3)
Now,
(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)
= (x+3)(x−2)(x−3)(x−4)(x−2)(x−4)(x−3)(x+3) = 1

(vi) p4−16p2−4p+4
Solution:

Ganita Manjari Class 9 Maths Chapter 4 End of Chapter Exercise Solutions

Exploring Algebraic Identities End of Chapter Exercise Solutions

Question 1.
Use suitable identities to find the following products:
(i) (-3x + 4)²
Solution:
(-3x + 4)²
We can expand this using the identity
(a + b)² = a² + 2ab + b²
(-3x + 4)² = (-3x)² + 2(-3x)(4) + 4²
= 9x² – 24x + 16

(ii) (2s + 7) (2s – 7)
Solution:
(2s + 7) (2s – 7)
We can use the identity
(a + b)(a – b) = a² – b²
(25 + 7) (25 – 7) = (25)² – (7)²
= 4s² – 49

(iii) (p² + 12)(p² – 12)
Solution:
(p² + 12)(p² – 12)
Using the identity: (a + b)(a – b) = a² – b²
(p² + 12)(p² – 12)
= (p²)² – (12)²
= p⁴ – 14

(iv) (2n + 7) (2n – 7)
Solution:
(2n + 1)(2n – 1)
Using the identity: (a + b)(a – b) = a² – b²
(2n + 7) (2n – 7) = (2n)² – 7²
= 4n² – 49

(v) (s – 2t) (s² + 2st + 4t²)
Solution:
(s – 2t) (s² + 2st + 4t²)
Here, we use the identity
(x – y)(x² + xy + y²) = x³ – y³
(s – 2t) (s² + 2st + 4t²) = s² – (2t)²
= s³ – 8t³

(vi) (1/2r – 4r)²
Solution:
(1/2r – 4r)²
We simplify this by using the identity
(a – b)² = a² – 2ab + b²
(1/2r)² – 2(1/2r)(4r) + (4r)²
= 1/4r2 – 4 + 16r²

(vii) (-3m + 4k – l)²
Solution:
(-3m + 4k – l)²
We can expand using the identity:
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
(-3m + 4k – l)² = (—3m)² + (4k)² + (-l)² + 2(-3m)(4k) + 2(-3m)(-l) + 2(4k)(-l)
= 9m² + 16k² +12 — 24mk + 6ml – 8kl

(viii) (x – 1/3)³
Solution:
(x – 1/3)³
To expand this, we use the identity:
(a – b)³ = a³ – 3a²b + 3ab² – b³
= a³ – b³ – 3ab(a – b)
(x – 1/3)³ = x³ – 1/27y³ – 3x x 1/3y(x – 1/3y)
= x³ – 1/27y – x²y + x³

(ix) (7/2 k – 2/3 m)³
Solution:
To expand this, we use the identity:
(a – b)³ = a³ – 3a²b + 3ab² – b³

Question 2.
Find the values using suitable identities:
(i) 17 × 21
Solution:
17 × 21
Here, 17 = 20 – 3 and 21 = 20 + 1, so:
17 × 21 = (20 – 3)(20 + 1)
Using the identity (x + a)(x + b)
= x² + (a + h)x + ab
= 20² + (-3 + 1)20 + (-3)(1)
= 400 – 40 – 3
= 357

(ii) 104 × 96
Solution:
104 × 96
Here, 104 = 100 + 4 and 96 = 100 – 4,
So, 104 × 96 = (100 + 4)(100 – 4)
Using the identity (a + b)(a – b) = a² – b²
= 100² – 4²
= 10000 – 16
= 9984

(iii) 24 × 16
Solution:
24 × 16
Here, 24 = 20 + 4 and 16 = 20 – 4
So, 24 × 16 = (20 + 4)(20 – 4)
Using the identity (a + b)(a – b) = a² – b²
= 20² – 4²
= 400- 16
= 384

(iv) 147³
Solution:
147³
We can use the identity for cubes:
(a – b)³ = a³ – 3a2b + 3 ab2 – b³
Here, 147 = 150 – 3,
So, 147³ = (150 – 3)³
= 150³ – 3 × 150² × 3 + 3 × 150 × 3² – 3³
Simplifying:
= 3375000 – 202500 + 4050 – 27
= 3176523

(v) 199³
Solution:
199³
We can use the identity for cubes:
(a – b)³= a³ – 3a²b + 3ab² – b³
Here, 199 = 200- 1,
So, 199³ = (200 – 1)³
= 2003 – 3 × 200² × 1 + 3 × 200 × 1² – 1³
= 8000000 – 120000 + 600 – 1
= 7880599

(vi) 127³
Solution:
127³
We can use the identity for cubes:
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, 127 = 130 – 3,
So, 127³ = (130 – 3)³
= 130³ – 3 × 130² × 3 + 3 × 130 × 3² – 3³
= 2197000 – 152100 + 3510 – 27
= 2048383

(vii) (–107)³
Solution:
(–107)³

(–100 –7)³
We can use the identity for cubes:
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here, –107 =–100 – 7,
So, (–107)³ = (–100 – 7)³
= (–100)³ – 3 × (–100)² × 7 + 3 × (–100) × 7² – 7³
= –1000000 – 210000 + (–14700) – 343
= –1225043

(viii) (–299)³
Solution:
(–299)³

(–300 +1)³
We can use the identity for cubes:
(a + b)³ = a³ + 3a²b + 3 ab² + b³
Here, –299 = –300 + 1,
So, (–299)³ = (–300 + 1)³
= (–300)³ + 3 × (–300)² × 1 + 3 × (–300) × 1²+ 1³
= –27000000 + 270000 – 900 + 1
= –26730899

Q. 3. Factor the following algebraic expressions:
(i) 4y² + 1 + 116y²

(ii) 9m² – 125n²

(iii) 27b³ – 164b³

(iv) x² + 5x6+16

(v) 27u³ – 1/125−27u²/5+9u/25

(vi) 64y³ – 1125z³

(vii) p³ + 27q³ + r³ – 9pqr

(viii) 9m² – 12m + 4

(ix) 9x³ – 83 y³ + z33 + 6xyz

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy

27u³ – 1/216−9u²/2+u/4

Solutions 

(i) 4y² + 1 + 116y²
Solution: aa
4y² + 1 + 116y²
= (2y)² + 2 × (2y) × (14y)+(14y)2
= (2y + 12y)²
[∵ (a + b)² = a² + 2ab + b²]

(ii) 9m² – 125n²
Solution:
9m² – 125n²
= (3m)² – (15n)²
= (3m + 15n)(3m – 15n)
[∵ a² – b² = (a + b)(a – b)]

(iii) 27b³ – 164b³
Solution:

(iv) x² + 5x6+16
Solution:

(v) 27u³ – 1/125−27u²/5+9u/25
Solution:

(vi) 64y³ – 1125z³
Solution:

(vii) p³ + 27q³ + r³ – 9pqr
Solution:
p³ + 27q³ + r³ – 9pqr
We have, x³ + y³ + z³ – 3xyz
= (x + y + z)(x³ + y³ + z³ – xy – xz – yz)
= (p + 3q + r)(p³ + 9q³ + r³ – 3pq – 3qr – pr)

(viii) 9m² – 12m + 4
Solution:
9m² – 12m + 4
= (3m)² – 2(3m)(2) + (2)²
= (3m – 2)²
[∵ (a – b)² = a² – 2ab + b²]

(ix) 9x³ – 83 y³ + z33 + 6xyz
Solution:
9x³ – 83 y³ + z33 + 6xyz
= 13(27x³ – 8y³ + z³ + 18xyz)
= 13[(3x)³ + (-2y)³ + (z)³ – 3(3x)(-2y)(z)]
= 13(3x – 2y +z)(9x³ + 4y³ + z³ + 6xy + 2yz – 3xz)
[∵ a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy
Solution:
4x² + 9y² + 36z² + 12xz + 36yz + 24xy
= (2x)² + (3y)² + (6z)² + 2(2x)(3y) + 1(3y)(6z) + 2(2x)(6z)
= (2x + 3y + 6z)²
[∵ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca)

(xi) 27u³ – 1/216−9u²/2+u/4
Solution:
27u³ – 1/216−9u²/2+u/4
= (3u)³ – (16)³ – 3(3u)²(16) + 3(3u)(16)²
= (3u – 16)³
[∵ (a – b)³ = a³ – b³ – 3a²b + 3ab²]

Question 4.
Simplify the following:
(i) (4x²+4x+1)/(4x²−1)
(ii) 9(3a³−24b³)/(9a²−36b²)
(iii) (s³+125t³) /(s²−2st−35t²)
Note: Assume that the denominators are not equal to 0.
Solution:

Question 5.
Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) 25a² – 30ab + 9b² (ii) 36s² – 49t²

(i) 25a² – 30ab + 9b²

Solution:
25a² – 30ab + 9b²
= (5a)² – 2(5a)(3b) + (3b)²
= (5a – 3b)²
[∵ (a – b) = a² – 2ab + b²]
= (5a – 3b)(5a – 3b)
Area of a rectangle = length × breadth
Possible dimensions:
Length = (5a – 3b) units
Breadth = (5a – 3b) units

(ii) 36s² – 49t²
Solution:
36s² – 49t²
= (6s)² – (7t)²
= (6s + 7t)(6s – 7t)
[∵ a² – b² = (a + b) (a – b)]
Area of a rectangle = length × breadth
Possible dimensions:
Length = (6s + 7t) units
Breadth = (6s – 7t) units

Question 6.
Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) 6a² – 24b^2.    (ii) 3ps² – 15ps +12p

(i) 6a² – 24b^2.

Solution:
6a² – 24b²
= 6(a² – 4b²)
= 6[a² -(2b)²]
= 6(a + 2 b)(a – 2b) [∵ a² – b² = (a + b)(a – b)]
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 6 units
Breadth = (a + 2b) units
Height = (a – 2b) units

(ii) 3ps² – 15ps +12p
Solution:
3ps² – 15ps + 12p
= 3p(s² – 5s + 4)
= 3p(s² – 4s – s + 4)
= 3p[s(s – 4) – 1(s – 4)]
= 3p(s – 1)(s – 4)
Volume of a cuboid = length × breadth × height
Possible dimensions:
Length = 3p units
Breadth = (s – 1) units
Height = (s – 4) units

Q. 7. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.
Solution:

Area of the playground = 40 m × 40 m
= (40)² m²
Area of the playground along with the path
= (40 + 2s)m × (40 + 2s)m
= (40 + 2s)² m²
Area of the path = (40 + 2s)² – (40)²
= (40 + 2s + 40)(40 + 2s-40) [v a² – b² = (a + b)(a – b)]
= (80 + 2s)(2s)
= 160s + 4s²
So, the area of the path is (4s² + 160s) m².

Q. 8. If a number plus its reciprocal equals 10/3, find the number.
Solution:
Let the number be x
According to the statement, we have
x + 1/x=10/3
On multiplying both sides by 3x, we get
⇒ 3x² + 3 = 10x
⇒ 3x² – 10x + 3 = 0
Now, splitting the middle term, we get
⇒ 3x² – 9x – x + 3 = 0
⇒ 3x(x – 3) – 1(x – 3) = 0
⇒ (3x – 1) (x – 3) = 0
⇒ 3x – 1 = 0 or x – 3 = 0
⇒ x = 1/3 or x = 3
Hence, the required numbers are 1/3 and 3.

Q.  9. A rectangular pool has area 2a² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.
Solution:
Area of the pool = 2x² + 7x + 3 square hastas
= 2x² + 6x + x + 3 square hastas
= 2x(x + 3) + 10c+ 3) square hastas
= (2x + 1)(x + 3) square hastas
Area of the pool = length × breadth
Width of the pool = (2x + 1) hastas (given)
Therefore, length of the pool = (x + 3) hastas

Q. 10. If both x – 2 and x – 1/2 are factors of px² + 5x + r, show 2 that p = r.
Solution:
px² + 5x + r
Factors = x — 2, x – 1/2
⇒ x = 2 or 1/2
For x = 2,
p(2)² + 5(2) + r = 0
4p + 10 + r = 0 …(1)

For x = 1/2
p(12)² + 5(12) + r = 0
p4+52 + r = 0 …….(2)

Equating (1) and (2), we get

p = -2
Substituting p = -2 in equation (1), we get
4(-2) + 10 + r = 0
⇒ -8+ 10 + r = 0
⇒ 2 + r = 0
⇒ r = -2
∴ p = r = – 2
Hence, p = r

Q. 11. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ – 3abc = – 25.
Solution:
We know that
a³ + b³ + c³ – 3abc
= (a + b + c)(a² + b² + c² – ab – bc – ca)
= (a + b + c) [a² + b² + c² – (ab + be + ca)]
= 5(a² + b² + c² – (ab + bc + ca))
= 5(a² + b² + c² – 10) …(1)
Now, a + b + c = 5
Squaring both sides, we get
(a + b + c)² = 5²
⇒ a² + b² + c² + 2(ab + bc + ca) = 25
⇒ a² + b² + c² + 2(10) = 25
⇒ a² + b² + c² = 25 – 20 = 5
Substituting a² + b² + c² = 5 in equation (1)
5(5 – 10) = 5(-5) = — 25
Hence proved.

Q.  12. By factoring the expression, check that n³ – n is, always divisible by 6 for all natural numbers n. Give reasons.
Solution:
Given, n³ – n
Now, factorising the expression, 

we get n³ – n = n(n² – 1)
Also, using the identity a² – b² = (a – b) (a + b), we get
n(n² – 1) = n(n – 1) (n + 1)
Now, n – 1, n, n + 1 are three consecutive natural numbers, and among any three consecutive numbers, one is divisible by 2, and also one is divisible by 3.
Hence, the product (n – 1) (n + 1) is divisible by 2 × 3 = 6.
Therefore, n³ – n is always divisible by 6 for all natural numbers n.

Q.  13. Find the value of
(i) x³ + y³ – 12xy + 64, when x + y = – 4

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6

(i) x³ + y³ – 12xy + 64, when x + y = – 4

Solution:

x³ + y³ – 12xy – 64
Using the identity: (a + b)³ = a³ + b³ + 3ab(a + b)
(x + y)³ = x³ + y³ + 3xy(x + y)
⇒ (-4)³ = x³ + y³ + 3xy(—4)
[∵ x +y = —4]
⇒ -64 = x³ + y³ – 12xy
⇒ x³ + y³ — 12xy + 64 = 0

(ii) x³ – 8y³ – 36xy – 216, when x = 2y + 6
Solution:
We are given, x = 2y + 6
⇒ x – 2y = 6
Now, x³ – 8y³ – 36xy – 216
Using the identity: (a – b)³ = a³ – b³ – 3ab(a – b)
(x – 2y)³ = x³ – (2y)³ – 3x(2y)(x – 2y)
(6)³ = x³ – 8y³ – 6xy(6)
216 = x³ – 8y³ – 36xy
⇒ x³ – 8y³ – 36xy – 216 = 0