# SELINA Solutions for Class 9 Maths Chapter 4 - Expansion

Access TopperLearning’s free Selina Solutions for ICSE Class 9 Mathematics Chapter 4 Expansion to learn about finding the square of a given algebraic expression using expansion. Also, go through the solutions to understand expansion of cubes. Highly-experienced Maths experts have created the answers for textbook exercises in a step-wise format to help it easy for students to understand and learn.

With thorough revision using Selina textbook solutions, you will learn to evaluate identities using the right methods. If you need explanation on the basics of expansion, check our ICSE Class 9 Maths concept videos and other chapter resources.

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## Chapter 4 - Expansion Exercise Ex. 4(A)

Question 1

Find the square of:

(i) 2a + b

(ii) 3a + 7b

(iii) 3a - 4b

(iv)

Solution 1

Question 2

Use identities to evaluate:

(i) (101)2

(ii) (502)2

(iii) (97)2

(iv) (998)2

Solution 2

Question 3

Evalute:

(i)

(ii)

Solution 3

(i)

(ii)

Question 4

Evaluate:

(i)

(ii) (4a +3b)2 - (4a - 3b)2 + 48ab.

Solution 4

(i)Consider the given expression:

(ii)Consider the given expression:

Question 5

If a + b = 7 and ab = 10; find a - b.

Solution 5

Question 6

If a -b = 7 and ab = 18; find a + b.

Solution 6

Question 7

If x + y = and xy = ; find:

(i) x - y

(ii) x2- y2

Solution 7

(i)

(ii)

Question 8

If a - b = 0.9 and ab = 0.36; find:

(i) a + b

(ii) a2 - b2.

Solution 8

(i)

(ii)

Question 9

If a - b = 4 and a + b = 6; find

(i) a2 + b2

(ii) ab

Solution 9

(i)

(ii)

Question 10

If a + = 6 and  a ≠ 0 find :

(i)

(ii)

Solution 10

(i)

(ii)

Question 11

If a - = 8 and a ≠0, find :

(i)

(ii)

Solution 11

(i)

(ii)

Question 12

If a2 - 3a + 1 = 0, and a≠ 0; find:

(i)

(ii)

Solution 12

(i)

(ii)

Question 13

If a2 - 5a - 1 = 0 and a ≠ 0; find:

(i)

(ii)

(iii)

Solution 13

(i)

(ii)

(iii)

Question 14

If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2.

Solution 14

Question 15

The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.

Solution 15

Given x is 2 more than y, so x = y + 2

Sum of squares of x and y is 34, so x2 + y2 = 34.

Replace x = y + 2 in the above equation and solve for y.

We get (y + 2)2 + y2 = 34

2y2 + 4y - 30 = 0

y2 + 2y - 15 = 0

(y + 5)(y - 3) = 0

So y = -5 or 3

For y = -5, x =-3

For y = 3, x = 5

Product of x and y is 15 in both cases.

Question 16

The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.

Solution 16

Let the two positive numbers be a and b.

Given difference between them is 5 and sum of squares is 73.

So a - b = 5, a2 + b2 = 73

Squaring on both sides gives

(a - b)2 = 52

a2 + b2 - 2ab = 25

But a2 + b2 = 73

So 2ab = 73 - 25 = 48

ab = 24

So, the product of numbers is 24.

## Chapter 4 - Expansion Exercise Ex. 4(B)

Question 1

Find the cube of :

(i) 3a- 2b

(ii) 5a + 3b

(iii)

(iv)

Solution 1

(i)

(ii)

(iii)

(iv)

Question 2

If a2 + = 47 and ≠ 0 find:

(i)

(ii)

Solution 2

(i)

(ii)

Question 3

If a2 + = 18; a ≠ 0 find:

(i)

(ii)

Solution 3

(i)

(ii)

Question 4

If a + = p and a ≠ 0 ; then show that:

Solution 4

Question 5

If a + 2b = 5; then show that:

a3 + 8b3 + 30ab = 125.

Solution 5

Question 6

If  and a ≠ 0 ; then show: a3 +

Solution 6

Question 7

If a + 2b + c = 0; then show that:

a3 + 8b3 + c3 = 6abc.

Solution 7

Question 8

Use property to evaluate:

(i) 133 + (-8)3 + (-5)3

(ii)73 + 33 + (-10)3

(iii) 93 - 53 - 43

(iv) 383 + (-26)3 + (-12)3

Solution 8

Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc

(i) a = 13, b = -8 and c = -5

133 + (-8)3 + (-5)3 = 3(13)(-8)(-5) = 1560

(ii) a = 7, b = 3, c = -10

73 + 33 + (-10)3 = 3(7)(3)(-10) = -630

(iii)a = 9, b = -5, c = -4

93 - 53 - 43 = 93 + (-5)3 + (-4)3 = 3(9)(-5)(-4) = 540

(iv) a = 38, b = -26, c = -12

383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568

Question 9

Solution 9

(i)

(ii)

Question 10

If and a - = 4; find:

(i)

(ii)

(iii)

Solution 10

(i)

(ii)

(iii)

Question 11

If and  x + = 2; then show that:

Solution 11

Thus from equations (1), (2) and (3), we have

Question 12

If 2x - 3y = 10 and xy = 16; find the value of 8x3 - 27y3.

Solution 12

Given that 2x - 3y = 10, xy = 16

Question 13

Expand :

(i)  (3x + 5y + 2z) (3x - 5y + 2z)

(ii)  (3x - 5y - 2z) (3x - 5y + 2z)

Solution 13

(i)

(3x + 5y + 2z) (3x - 5y + 2z)

= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}

= (3x + 2z)2 - (5y)2

{since (a + b) (a - b) = a2 - b2}

= 9x2 + 4z2 + 2 × 3x × 2z - 25y2

= 9x2 + 4z2 + 12xz - 25y2

= 9x2 + 4z- 25y2 + 12xz

(ii)

(3x - 5y - 2z) (3x - 5y + 2z)

= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}

= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}

= 9x2 + 25y2 - 2 × 3x × 5y - 4z2

= 9x2 + 25y2- 30xy - 4z2

= 9x2 +25y2 - 4z2 - 30xy

Question 14

The sum of two numbers is 9 and their product is 20. Find the sum of their

(i) Squares (ii) Cubes

Solution 14

Given sum of two numbers is 9 and their product is 20.

Let the numbers be a and b.

a + b = 9

ab = 20

Squaring on both sides gives

(a+b)2 = 92

a2 + b2 + 2ab = 81

a2 + b2 + 40 = 81

So sum of squares is 81 - 40 = 41

Cubing on both sides gives

(a + b)3 = 93

a3 + b3 + 3ab(a + b) = 729

a3 + b3 + 60(9) = 729

a3 + b3 = 729 - 540 = 189

So the sum of cubes is 189.

Question 15

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:

(i) Sum of these numbers

(ii) Difference of their cubes

(iii) Sum of their cubes.

Solution 15

Given x - y = 5 and xy = 24 (x>y)

(x + y)2 = (x - y)2 + 4xy = 25 + 96 = 121

So, x + y = 11; sum of these numbers is 11.

Cubing on both sides gives

(x - y)3 = 53

x3 - y3 - 3xy(x - y) = 125

x3 - y3 - 72(5) = 125

x3 - y3= 125 + 360 = 485

So, difference of their cubes is 485.

Cubing both sides, we get

(x + y)3 = 113

x3 + y3 + 3xy(x + y) = 1331

x3 + y3 = 1331 - 72(11) = 1331 - 792 = 539

So, sum of their cubes is 539.

Question 16

If 4x2 + y2 = a and xy = b, find the value of 2x + y.

Solution 16

xy = b ….(i)

4x2 + y2 = a ….(ii)

Now, (2x + y)2 = (2x)2 + 4xy + y2

= 4x2 + y2 + 4xy

= a + 4b ….[From (i) and (ii)]

## Chapter 4 - Expansion Exercise Ex. 4(C)

Question 1

Expand:

(i) (x + 8) (x + 10)

(ii) (x + 8) (x - 10)

(iii) (x - 8) (x + 10)

(iv) (x - 8) (x - 10)

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.

Solution 4

Question 5

If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.

Solution 5

Question 6

If a + b + c = p and ab + bc + ca = q; find a2 + b2 + c2.

Solution 6

Question 7

If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.

Solution 7

Question 8

If x+ y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.

Solution 8

## Chapter 4 - Expansion Exercise Ex. 4(D)

Question 1

If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz; evaluate:

Solution 1

Given that x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0

Therefore, x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y

Now

Question 2

If a + = m and ≠ 0 ; find in terms of 'm'; the value of :

(i)

(ii)

Solution 2

(i)

(ii)

Question 3

In the expansion of (2x2 - 8) (x - 4)2; find the value of

(i) coefficient of x3

(ii) coefficient of x2

(iii) constant term.

Solution 3

Question 4

If x > 0 and find: .

Solution 4

Given that

Question 5

If 2(x2 + 1) = 5x, find :

(i) (ii)

Solution 5

(i)

2(x2 + 1} = 5x

Dividing by x, we have

(ii)

Question 6

If a2 + b2 = 34 and ab = 12; find:

(i) 3(a + b)2 + 5(a - b)2

(ii) 7(a - b)2 - 2(a + b)2

Solution 6

a2 + b2 = 34, ab= 12

(a + b)2 = a2 + b2 + 2ab

= 34 + 2 x 12 = 34 + 24 = 58

(a - b)2 = a2 + b2 - 2ab

= 34 - 2 x 12 = 34- 24 = 10

(i) 3(a + b)2 + 5(a - b)2

= 3 x 58 + 5 x 10 = 174 + 50

= 224

(ii) 7(a - b)2 - 2(a + b)2

= 7 x 10 - 2 x 58 = 70 - 116 = -46

Question 7

If 3x -  and x ≠ 0 find : .

Solution 7

Given 3x -

We need to find

Question 8

If x2 + = 7 and  x ≠ 0; find the value of:

.

Solution 8

Given that

We need to find the value of

Consider the given equation:

Question 9

If x = and x ≠ 5 find .

Solution 9

By cross multiplication,

=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x

Dividing both sides by x,

Question 10

If x =  and x ≠ 5 find .

Solution 10

By cross multiplication,

=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x

Dividing both sides by x,

Question 11

If 3a + 5b + 4c = 0, show that:

27a3 + 125b3 + 64c3 = 180 abc

Solution 11

Given that 3a + 5b + 4c = 0

3a + 5b = -4c

Cubing both sides,

(3a + 5b)3 = (-4c)3

=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3

=>27a3 + 125b3 + 45ab x (-4c) = -64c3

=>27a3 + 125b3 - 180abc = -64c3

=>27a3 + 125b3 + 64c3 = 180abc

Hence proved.

Question 12

The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.

Solution 12

Let a, b be the two numbers

.'. a + b = 7 and a3 + b3 = 133

(a + b)3 = a3 + b3 + 3ab (a + b)

=> (7)3 = 133 + 3ab (7)

=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210

=> 21ab = 210 => ab= 2I

Now a2 + b2 = (a + b)2 - 2ab

=72 - 2 x 10 = 49 - 20 = 29

Question 13

In each of the following, find the value of 'a':

(i) 4x2 + ax + 9 = (2x + 3)2

(ii) 4x2 + ax + 9 = (2x - 3)2

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Solution 13

(i) 4x2 + ax + 9 = (2x + 3)2

Comparing coefficients of x terms, we get

ax = 12x

so, a = 12

(ii) 4x2 + ax + 9 = (2x - 3)2

Comparing coefficients of x terms, we get

ax = -12x

so, a = -12

(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2

Comparing coefficients of x terms, we get

(7a - 5)x = 30x

7a - 5 = 30

7a = 35

a = 5

Question 14

If

(i)  (ii)

Solution 14

Given

Question 15

The difference between two positive numbers is 4 and the difference between their cubes is 316.

Find:

(i) Their product

(ii) The sum of their squares

Solution 15

Given difference between two positive numbers is 4 and difference between their cubes is 316.

Let the positive numbers be a and b

a - b = 4

a3 - b3 = 316

Cubing both sides,

(a - b)3 = 64

a3 - b3 - 3ab(a - b) = 64

Given a3 - b3 = 316

So 316 - 64 = 3ab(4)

252 = 12ab

So ab = 21; product of numbers is 21

Squaring both sides, we get

(a - b)2 = 16

a2 + b2 - 2ab = 16

a2 + b2 = 16 + 42 = 58

Sum of their squares is 58.

## Chapter 4 - Expansion Exercise Ex. 4(E)

Question 1

Simplify:

(i) (x + 6)(x + 4)(x - 2)

(ii) (x - 6)(x - 4)(x + 2)

(iii) (x - 6)(x - 4)(x - 2)

(iv) (x + 6)(x - 4)(x - 2)

Solution 1

Using identity:

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

(i) (x + 6)(x + 4)(x - 2)

= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)

= x3 + 8x2 + (24 - 8 - 12)x - 48

= x3 + 8x2 + 4x - 48

(ii) (x - 6)(x - 4)(x + 2)

= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2

= x3 - 8x2 + (24 - 8 - 12)x + 48

= x3 - 8x2 + 4x + 48

(iii) (x - 6)(x - 4)(x - 2)

= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)

= x3 - 12x2 + (24 + 8 + 12)x - 48

= x3 - 12x2 + 44x - 48

(iv) (x + 6)(x - 4)(x - 2)

= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)

= x3 - 0x2 + (-24 + 8 - 12)x + 48

= x3 - 28x + 48

Question 2

Solution 2

Question 3

Using suitable identity, evaluate

(i) (104)3

(ii) (97)3

Solution 3

Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)

(i) (104)3 = (100 + 4)3

= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)

= 1000000 + 64 + 1200 × 104

= 1000000 + 64 + 124800

= 1124864

(ii) (97)3 = (100 - 3)3

= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)

= 1000000 - 27 - 900 × 97

= 1000000 - 27 - 87300

= 912673

Question 4

Solution 4

Question 5

Solution 5

Question 6

If a - 2b + 3c = 0; state the value of a3 - 8b3 + 27c3.

Solution 6

a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

Since a - 2b + 3c = 0, we have

a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3

= 3(a)( -2b)(3c)

= -18abc

Question 7

If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.

Solution 7

x + 5y = 10

(x + 5y)3 = 103

x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000

x3 + (5y)3 + 3(x)(5y)(10) = 1000

= x3 + (5y)3 + 150xy = 1000

= x3 + (5y)3 + 150xy - 1000 = 0

Question 8

Solution 8

Question 9

If a + b = 11 and a2 + b2 = 65; find a3 + b3.

Solution 9

Question 10

Prove that:

x2+ y2 + z2 - xy - yz - zx  is always positive.

Solution 10

x2 + y2 + z2 - xy - yz - zx

= 2(x2 + y2 + z2 - xy - yz - zx)

= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx

= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx

= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)

= (x - y)2 + (z - x)2 + (y - z)2

Since square of any number is positive, the given equation is always positive.

Question 11

Find:

(i) (a + b)(a + b)

(ii) (a + b)(a + b)(a + b)

(iii) (a - b)(a - b)(a - b) by using the result of part (ii)

Solution 11

(i) (a + b)(a + b) = (a + b)2

= a × a + a × b + b × a + b × b

= a2 + ab + ab + b2

= a2 + b2 + 2ab

(ii) (a + b)(a + b)(a + b)

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

= (a2 + ab + ab + b2)(a + b)

= (a2 + b2 + 2ab)(a + b)

= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b

= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2

= a3 + b3 + 3a2b + 3ab2

(iii) (a - b)(a - b)(a - b)

In result (ii), replacing b by -b, we get

(a - b)(a - b)(a - b)

= a3 + (-b)3 + 3a2(-b) + 3a(-b)2

= a3 - b3 - 3a2b + 3ab2

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