 # ML Aggarwal Solutions for Class 9 Maths Chapter 3 - Expansions

ML Aggarwal Solutions for Class 9 Maths Chapter 3 – Expansions are provided here to help students prepare and excel in their exams. This chapter mainly deals with problems based on expansions. Experts tutors have formulated the solutions in a step by step manner for students to grasp the concepts easily. From the exam point of view, solving the problems on a regular basis, students can improve their conceptual knowledge. The solutions contain brief step wise explanations, which are purely based on the latest syllabus of ICSE board. Students can cross check the answers with the solutions designed by the experts and understand the other ways of solving problems effortlessly. To know more about these concepts, students can access ML Aggarwal Solutions pdf, from the links which are provided below and start practising offline to secure good marks in the board exams.

Chapter 3 – Expansions contain chapter test and the ML Aggarwal Class 9 Solutions present in this page provide solutions to questions related to each topic present in this chapter.

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Chapter Test

1. Find the expansions of the following :
(i) (2x + 3y + 5) (2x + 3y – 5)
(ii) (6 – 4a -7b)2
(iii) (7 – 3xy)3
(iv) (x + y + 2)3

Solution:

(i) (2x + 3y + 5) (2x + 3y – 5)

Let us simplify the expression, we get

(2x + 3y + 5) (2x + 3y – 5) = [(2x + 3y) + 5] [(2x – 3y) – 5]

By using the formula, (a)2 – (b)2 = [(a + b) (a – b)]
= (2x + 3y)2 – (5)2

= (2x)2 + (3y) 2 + 2 × 2x × 3y – 5 × 5

= 4x2 + 9y2 + 12xy – 25

(ii) (6 – 4a – 7 b)2

Let us simplify the expression, we get

(6 – 4a – 7 b)2 = [ 6 + (- 4a) + (-7b)]2

= (6)2 + (- 4a)2 + (- 7b)2 + 2 (6) (- 4a) + 2 (- 4a) (-7b) + 2 (-7b) (6)

= 36 + 16a2 + 49b2 – 48a + 56ab – 84b

(iii) (7 – 3xy)3

Let us simplify the expression

By using the formula, we get

(7 – 3xy)3 = (7)3 – (3xy)3 – 3 (7) (3xy) (7 – 3xy)

= 343 – 27x3y3 – 63xy (7 – 3xy)

= 343 – 27x3y3 – 441xy + 189x2y2

(iv) (x + y + 2)3

Let us simplify the expression

By using the formula, we get

(x + y + 2 )3 = [(x + y) + 2]3

= (x + y)3 + (2)3 + 3 (x + y) (2) (x + y + 2)

= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x + y) [(x + y) + 2]
= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x + y)2 +12(x + y)

= x3 + y3 + 3x2y + 3xy2 + 8 + 6 (x2 + y2 + 2xy) + 12x + 12y = x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12xy + 12x + 12y

= x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12x + 12y + 12xy

2. Simplify: (x – 2) (x + 2) (x2 + 4) (x4 + 16)
Solution:

Let us simplify the expression, we get

(x – 2) (x + 2) (x4 + 4) (x4 +16) = (x2 – 4) (x4 + 4) (x4 + 16)

= [(x2)2 – (4)2] (x4 + 16)

= (x4 – 16) (x4 + 16)

= (x4)2 – (16)2

= x8 – 256

3. Evaluate 1002 × 998 by using a special product.
Solution:

Let us simplify the expression, we get

1002 × 998 = (1000 + 2) (1000 – 2)

= (1000)2 – (2)2

= 1000000 – 4

= 999996

4. If a + 2b + 3c = 0, Prove that a3 + 8b3 + 27c3 = 18 abc
Solution:

Given:

a + 2b + 3c = 0, a + 2b = – 3c

Let us cube on both the sides, we get

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

a3 + (2b)3 + 3(a) (2b) (a + 2b) = -27c3

a3 + 8b3 + 6ab (– 3c) = – 27c3

a3 + 8b3 – 18abc = -27c3

a3 + 8b3 + 27c3 = 18abc

Hence proved.

5. If 2x = 3y – 5, then find the value of 8x3 – 27y3 + 90xy + 125.
Solution:

Given:

2x = 3y – 5

2x – 3y = -5

Now, let us cube on both sides, we get

(2x – 3y)3 = (-5)3

(2x)3 – (3y)3 – 3 × 2x × 3y (2x – 3y) = -125

8x3 – 27y3 – 18xy (2x – 3y) = -125

Now, substitute the value of 2x – 3y = -5

8x3 – 27y3 – 18xy (-5) = -125

8x3 – 27y3 + 90xy = -125

8x3 – 27y3 + 90xy + 125 = 0

6. If a2 – 1/a2 = 5, evaluate a4 + 1/a4

Solution:

It is given that,

a2 – 1/a2 = 5

So,

By using the formula, (a + b)2

[a2 – 1/a2]2 = a4 + 1/a4 – 2

[a2 – 1/a2]2 + 2 = a4 + 1/a4

Substitute the value of a2 – 1/a2 = 5, we get

52 + 2 = a4 + 1/a4

a4 + 1/a4 = 25 + 2

= 27

7. If a + 1/a = p and a – 1/a = q, Find the relation between p and q.

Solution:

It is given that,

a + 1/a = p and a – 1/a = q

so,

(a + 1/a)2 – (a – 1/a)2 = 4(a) (1/a)

= 4

By substituting the values, we get

p2 – q2 = 4

Hence the relation between p and q is that p2 – q2 = 4.

8. If (a2 + 1)/a = 4, find the value of 2a3 + 2/a3

Solution:

It is given that,

(a2 + 1)/a = 4

a2/a + 1/a = 4

a + 1/a = 4

So by multiplying the expression by 2a, we get

2a3 + 2/a3 = 2[a3 + 1/a3]

= 2 [(a + 1/a)3 – 3 (a) (1/a) (a + 1/a)]

= 2 [(4)3 – 3(4)]

= 2 [64 – 12]

= 2 (52)

= 104

9. If x = 1/(4 – x), find the value of

(i) x + 1/x

(ii) x3 + 1/x3

(iii) x6 + 1/x6

Solution:

It is given that,

x = 1/(4 – x)

So,

(i) x(4 – x) = 1

4x – x2 = 1

Now let us divide both sides by x, we get

4 – x = 1/x

4 = 1/x + x

1/x + x = 4

1/x + x = 4

(ii) x3 + 1/x3 = (x + 1/x)2 – 3(x + 1/x)

By substituting the values, we get

= (4)3 – 3(4)

= 64 – 12

= 52

(iii) x6 + 1/x6 = (x3 + 1/x3)2 – 2

= (52)2 – 2

= 2704 – 2

= 2702

10. If x – 1/x = 3 + 2√2, find the value of ¼ (x3 – 1/x3)

Solution:

It is given that,

x – 1/x = 3 + 2√2

So,

x3 – 1/x3 = (x – 1/x)3 + 3(x – 1/x)

= (3 + 2√2)3 + 3(3 + 2√2)

By using the formula, (a+b)3 = a3 + b3 + 3ab (a + b)

= (3)3 + (2√2)3 + 3 (3) (2√2) (3 + 2√2) + 3(3 + 2√2)

= 27 + 16√2 + 54√2 + 72 + 9 + 6√2

= 108 + 76√2

Hence,

¼ (x3 – 1/x3) = ¼ (108 + 76√2)

= 27 + 19√2

11. If x + 1/x = 3 1/3, find the value of x3 – 1/x3

Solution:

It is given that,

x + 1/x = 3 1/3

we know that,

(x – 1/x)2 = x2 + 1/x2 – 2

= x2 + 1/x2 + 2 – 4

= (x + 1/x)2 – 4

But x + 1/x = 3 1/3 = 10/3

So,

(x – 1/x)2 = (10/3)2 – 4

= 100/9 – 4

= (100 – 36)/9

= 64/9

x – 1/x = √(64/9)

= 8/3

Now,

x3 – 1/x3 = (x – 1/x)3 + 3 (x) (1/x) (x – 1/x)

= (8/3)3 + 3 (8/3)

= ((512/27) + 8)

= 728/27

= 26 26/27

12. If x = 2 – √3, then find the value of x3 – 1/x3

Solution:

It is given that,

x = 2 – √3

so,

1/x = 1/(2 – √3)

By rationalizing the denominator, we get

= [1(2 + √3)] / [(2 – √3) (2 + √3)]

= [(2 + √3)] / [(22) – (√3)2]

= [(2 + √3)] / [4 – 3]

= 2 + √3

Now,

x – 1/x = 2 – √3 – 2 – √3

= – 2√3

Let us cube on both sides, we get

(x – 1/x)3 = (-2√3)3

x3 – 1/x3 – 3 (x) (1/x) (x – 1/x) = 24√3

x3 – 1/x3 – 3(-2√3) = -24√3

x3 – 1/x3 + 6√3 = -24√3

x3 – 1/x3 = -24√3 – 6√3

= -30√3

Hence,

x3 – 1/x3 = -30√3

13. If the sum of two numbers is 11 and sum of their cubes is 737, find the sum of their squares.
Solution:

Let us consider x and y be two numbers

Then,

x + y = 11

x3 + y3 = 735 and x2 + y2 =?

Now,

x + y = 11

Let us cube on both the sides,

(x + y)3 = (11)3

x3 + y3 + 3xy (x + y) = 1331

737 + 3x × 11 = 1331

33xy = 1331 – 737

= 594

xy = 594/33

xy = 8

We know that, x + y = 11

By squaring on both sides, we get

(x + y)2 = (11)2

x2 + y2 + 2xy = 121 2 x2 + y2 + 2 × 18 = 121

x2 + y2 + 36 = 121

x2 + y2 = 121 – 36

= 85

Hence sum of the squares = 85

14. If a – b = 7 and a3 – b3 = 133, find:
(i) ab
(ii) a2 + b2
Solution:

It is given that,

a – b = 7

let us cube on both sides, we get

(i) (a – b)3 = (7)3

a3 + b3 – 3ab (a – b) = 343

133 – 3ab × 7 = 343

133 – 21ab = 343

– 21ab = 343 – 133 21ab

= 210

ab = -210/21

ab = -10

(ii) a2 + b2

Again a – b = 7

Let us square on both sides, we get

(a – b)2 = (7)2

a2 + b2 – 2ab = 49

a2 + b2 – 2 × (- 10) = 49

a2 + b2 + 20 = 49

a2 + b2 = 49 – 20

= 29

Hence, a2 + b2 = 29

15. Find the coefficient of x2 expansion of (x2 + x + 1)2 + (x2 – x + 1)2
Solution:

Given:

The expression, (x2 + x + 1)2 + (x2 – x + 1)2

(x2 + x + 1)2 + (x2 – x + 1)2 = [((x2 + 1) + x)2 + [(x2 + 1) – x)2]
= (x2 + 1)2 + x2 + 2 (x2 + 1) (x) + (x2 + 1)2 + x2 – 2 (x2 + 1) (x)

= (x2)2 + (1)2 + 2 × x2 × 1 + x2 + (x2)2 + 1 + 2 × x2 + 1 + x2

= x4 + 1 + 2x2 + x2 + x4 + 1 + 2x2 + x2

= 2x4 + 6x2 + 2

∴ Co-efficient of x2 is 6.