M 9104. Exploring Algebraic Identities
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.
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.
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.
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.
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?
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 and 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
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.
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
Think and Reflect
What if we replace 'b' by '–b' in (a + b)² = a² + 2ab + b² ?
[(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
Let us revisit the pattern we observed in
Example 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: Suppose we have to calculate 29².
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
(a – b)²
= a² – ab – b(a – b)
= a² – ab – ba + b²
= a² – 2ab + b²
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
(iii) 64p² + (32/3) × pq + (4/9) × q²
(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)
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
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
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
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
= (110 + 7)²
= (110)² + 2 × 110 × 7 + 7²
= 12100 + 1540 + 49
= 37200 + 49
= 37249
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
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
4. Is this an identity?
(a + b – c)² + (a – b + c)² + (a – b – c)² = 2a² + 2b² + 2c²
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?
2. Observe the two rows of figures below. They represent an algebraic identity.
Try to identify it.
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.
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.
2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.
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)
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 = (______) (______)
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
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
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) (4y ^ 2 - 20yz + 25z ^ 2)/(25z ^ 2 - 4y ^ 2)
(v) ((x ^ 2 + x - 6)(x ^ 2 - 7x + 12))/((x ^ 2 - 6x + 8)(x ^ 2 - 9))
(vi) (p ^ 4 - 16)/(p ^ 2 - 4p + 4)
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
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
3. Factor the following algebraic expressions:
(i) 4y ^ 2 + 1 + 1/(16y ^ 2)
(ii) 9m ^ 2 - 1/(25n ^ 2)
3. Factor the following algebraic expressions:
(i) 4y ^ 2 + 1 + 1/(16y ^ 2)
(ii) 9m ^ 2 - 1/(25n ^ 2)
(iii) 27b ^ 3 - 1/(64b ^ 3)
(iv) x ^ 2 + (5x)/6 + 1/6
(v) 27u ^ 3 - 1/125 - (27u ^ 2)/5 + (9u)/25
(vi) 64y ^ 3 + 1/125 * z ^ 3
(vii) p³+27q³+r3-9pqr
(viii) 9m ^ 2 - 12m + 4
8 23 (ix) 9x3 + +6xyz 3 3
(x) 4x ^ 2 + 9y ^ 2 + 36z ^ 2 + 12xz + 36yz + 24xy
(xi) 27u ^ 3 - 1/216 - (9u ^ 2)/2 + u/4
4. Simplify the following:
(i) (4x ^ 2 + 4x + 1)/(4x ^ 2 - 1)
(ii) (9(3a ^ 3 - 24b ^ 3))/(9a ^ 2 - 36b ^ 2)
(iii) (s ^ 3 + 125t ^ 3)/(s ^ 2 - 2st - 35t ^ 2)
Note: Assume that the denominators are not equal to 0.
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
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
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.
8. If a number plus its reciprocal equals 10/3 find the number.
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.
*10. If both x – 2 and x – 1/2 are factors of px² + 5x + r show that p = r
*11. If a + b + c = 5 and ab + bc + ca = 10 then prove that a³+b³+c³–3abc = –25.
*12. By factoring the expression, check that n³ – n is always divisible by 6 for all natural numbers n. Give reasons.
*13. Find the value of
(i) x³ + y³ – 12xy + 64 when x + y = – 4
(ii) x³ – 8y³ – 36xy – 216 when x = 2y + 6
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)
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