# ICSE Solutions for Selina Concise Chapter 4 Expansions (including Substitution) Class 9 Maths

### Exercise 4(A)

1.1. Find the square of : 2a + b

We Know that
(a + b)2 = a2  + b2  + 2ab
⇒ (2a + b)2  = 4a2  + b2  + 2×2a×b
= 4a2 + b2 + 4ab

1.2. Find the square of : 3a + 7b

We know that
(a + b)2 = a2 + b2 + 2ab
⇒ (3a + 7b)2 = 9a2 + 49b2 + 2×3a×7b
= 9a2 + 49b2 + 42ab

1.3. Find the square of : 3a – 4b

We know that
(a – b)2 = a2 + b2 – 2ab
⇒ (3a – 4b)2 = 9a2 + 16b2 – 2×3a×4b
= 9a2 + 16b2 – 24ab

1.4. Find the square of : 3a/2b – 2b/3a

We know that,
( a – b )2  = a2  + b2  – 2ab

2.1. Use identities to evaluate : (101)2

(101)2
⇒ (101)2 = (100 + 1)2
We know that,
(a + b)2 = a2 + b2 + 2ab
∴ (100 + 1)2 = 1002 + 12 + 2×100×1
= 10,000 + 1 + 200
= 10,201

2.2. Use identities to evaluate : (502)2

(502)2
⇒ (502)= (500 + 2)2
We know that,
(a + b)2 = a2 + b2 + 2ab
∴ (500 + 2)2 = 5002 + 22 + 2×500× 2
= 250000 + 4 + 2000
= 2,52,004

2.3. Use identities to evaluate : (97)2

(97)2
⇒ (97)2 = (100 – 3)2
We know that,
( a – b )2 = a2 + b2 – 2ab
∴ (100 – 3)2 = 1002 + 32 – 2×100×3
= 10000 + 9 – 600
= 9,409

2.4. Use identities to evaluate : (998)2

(998)2
⇒ (998)2 = (1000 – 2)2
We know that
(a – b)2 = a2 + b2 – 2ab
∴ (1000 – 2)2 = 10002 + 22 – 2×1000×2
= 1000000 + 4 – 4000
= 9,96,004

3.1. Evalute : [(7/8)x + (4/5)y]2

3.2. Evalute : [2x/7 – 7y/4]

4.1. Evaluate : [a+2b + 2b/a]2 – [a/2b – 2b/a]2 − 4

Consider the given expression :

4.2. Evaluate : (4a +3b)2 – (4a – 3b)2 + 48ab.

(4a +3b)2 – (4a – 3b)2 + 48ab.
Consider the given expression:
Let us expand the first term : (4a +3b)2
We know that,
(a + b)2 = a2 + b2 + 2ab
∴ (4a +3b)= (4a)2 + (3b)2 + 2×4a×3b
= 16a2 + 9b2 + 24ab ...(1)

Let us expand the second term : (4a – 3b)2
We know that,
(a + b)2 = a2 + b2 + 2ab
∴ (4a – 3b)= (4a)2 + (3b)2 – 2×4a×3b
= 16a2 + 9b2 – 24ab …(2)

Thus from (1) and (2), the given expression is
(4a +3b)2 – (4a – 3b)2 + 48ab
= 16a2 + 9b2 + 24ab – 16a2 – 9b2 + 24ab + 48ab
= 96ab

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

We know that,
(a + b)2 = a2 + 2ab + b2
and
(a – b)2 = a2 – 2ab + b2
Rewrite the above equation, we have
(a – b)2 = a2 + b2 – 2ab + 4ab
= (a + b)2 – 4ab …(1)
Given that a + b = 7; ab = 10
Substitute the values of (a + b) and (ab)
in equation (1), we have
(a – b)2 = (7)2 – 4(10) = 49 – 40 = 9
⇒ a – b = ±√9
⇒ a – b = ±3

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

We know that,
(a – b)2 = a2 – 2ab + b2
and
(a + b)2 = a2 + 2ab + b2
Rewrite the above equation, we have
(a + b)2 = a2 + b2 – 2ab + 4ab
= (a + b)2 + 4ab …(1)
Given that a – b = 7; ab = 18
Substitute the values of (a – b) and (ab)
in equation (1), we have
(a + b)2 = (7)2 + 4(18) = 49 + 72 = 121
⇒ a + b = ±√121
⇒ a + b = ±11

7. If x + y = 72 and xy=5/2 ; find : x – y and x2 – y2

We know that,
(x + y)2 = x2 + 2xy + y2
and
(x – y)2 = x2 – 2xy + y2
Rewrite the above equation, we have
(x – y)2 = x2 + y2 + 2xy – 4xy = (x + y)2 – 4xy …(1)

8. If a – b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 – b2

(i) We know that,
(a – b)2 = a2 – 2ab + b2
and
(a + b)2 = a2 + 2ab + b2
Rewrite the above equation, we have
(a + b)2 = a2 + b2 – 2ab + 4ab
= (a – b)2 + 4ab …(1)
Given that a – b = 0.9 ; ab = 0.36
Substitute the values of (a – b) and (ab)
in equation (1), we have
(a + b)2 = (0.9)2 + 4(0.36) = 0.81 + 1.44 = 2.25

⇒ a + b = ±√2.25
⇒ a + b = ±1.5 ...(2)

(ii) We know that,
a2 – b2 = (a + b)(a – b) ….(3)
From equation (2) we have,
a + b = ±1.5
Thus equation (3) becomes,
a2 – b2 = (±1.5)(0.9) [ given a – b = 0.9 ]
⇒ a2 – b2 = ±1.35

9. If a – b = 4 and a + b = 6; find
(i) a2 + b2
(ii) ab

(i) We know that,
(a – b)2 = a2 – 2ab + b2
Rewrite the above identity as,
a2 + b= (a – b) + 2ab ...(1)
Similarly, we know that,
(a + b)2 = a2 + 2ab + b2
Rewrite the above identity as,
a2 + b2 = (a + b)2 – 2ab ...(2)
Adding the equations (1) and (2), we have
2(a2 + b2) = (a – b)2 + 2ab + (a + b)2 – 2ab
⇒ 2(a2 + b2) = (a – b)2 + (a + b)2

⇒ (a+ b2) = 26 ...(4)

From equation (4), we have
a2 + b2 = 26
Consider the identity,
(a – b)2 = a2 + b2 – 2ab ….(5)
Substitute the value a – b = 4 and a2 + b2 = 26
in the above equation, we have
(4)2 = 26 – 2ab
⇒ 2ab = 26 – 16
⇒ 2ab = 10
⇒ ab = 5

10. If a + 1/a= 6 and a ≠ 0 find :

(i) a – 1/a

(ii) a2 – 1/a

We know that,
(a + b)2 = a2 + 2ab + b2
and
(a – b)2 = a2 – 2ab + b
Thus,

11. If  a – 1/a = 8 and a ≠ 0; find :

(i) a +1/a

(ii) a2 – 1/a2

We know that,
(a + b)2 = a2 + 2ab + b2
and
(a – b)2 = a2 – 2ab + b
Thus,

12. If a2 – 3a + 1 = 0, and a ≠ 0; find :

(i) a + 1/a

(ii)  a2 + 1/a2

(i) Consider the given equation
a2 – 3a + 1 = 0
Rewrite the given equation, we have

13. If a2 – 5a – 1 = 0 and a ≠ 0 ; find :
(i) a – 1/a
(ii) a + 1/a
(iii) a2 – 1/a2

(i) Consider the given equation
a2  – 5a – 1 = 0

Rewrite the given equation, we have
a2  – 1 = 5a

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

Given that (3x + 4y) = 16 and xy = 4
We need to find 9x2 + 16y2.
We know that
(a + b)2 = a2 + b2 + 2ab
Consider the square of 3x + 4y :
∴ (3x + 4y)2 = (3x)2 + (4y)2 + 2 x 3x x 4y
⇒ (3x + 4y)2 = 9x2 + 16y2 + 24xy …..(1)

Substitute the values of (3x + 4y) and xy
in the above equation (1), we have
(3x + 4y)2 = 9x2 + 16y2 + 24xy

⇒ (16)2 = 9x2 + 16y2 + 24(4)

⇒ 256 = 9x2 + 16y2 + 96

⇒ 9x2 + 16y2 = 160

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.

Given x is 2 more than y, so x = y + 2
Sum of squares of x and y is 34, so x+ y= 34.
Replace x = y + 2 in the above equation and solve for y.
We get (y + 2)+ y= 34

2y+ 4y – 30 = 0
⇒ y+ 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.

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

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, a+ b= 73
Squaring on both sides gives
(a – b)= 52
⇒ a+ b– 2ab = 25
but a+ b= 73
So, 2ab = 73 – 25 = 48
ab = 24
So, the product of numbers is 24.

### Exercise 4(B)

1.1. Find the cube of : 3a- 2b

(a – b)3 = a3 – 3ab(a – b) – b3
(3a – 2b)3 = (3a)3 – 3×3a×2b(3a – 2b) – (2b)3
= 27a3 – 18ab(3a – 2b) – 8b3
= 27a3 – 54a2b + 36ab2 – 8b3

1.2. Find the cube of : 5a + 3b

(a + b)3 = a3 + 3ab(a + b) + b3
(5a + 3b)3 = (5a)3 + 3×5a×3b( 5a + 3b) + (3b)3
= 125a3 + 45ab(5a + 3b) + 27b3
= 125a3 + 225a2b + 135ab2 + 27b

1.3. Find the cube of : 2a + 1/2a ; (a ≠ 0)

(a + b)3  = a3  + 3ab(a + b) + b3

1.4. Find the cube of : 3a – 1/a ; (a ≠ 0)

( a – b )3 = a3 – 3ab( a – b ) – b3

2. If a2 + 1/a2 = 47 and a ≠ 0 ; find :

(i) a + 1/a

(ii) a + 1/a3

3. If a2 + 1/a2 = 18 and a ≠ 0; find :

(i) a – 1/a

(ii) a3 – 1/a3

4. If a + 1/a = p and a ≠ 0; then show that :

a3 + 1/a3 = p(p2 – 3)

5. If a + 2b = 5; then show that : a3 + 8b3 + 30ab = 125.

Given that a + 2b = 5 ;
We need to find a3 + 8b3 + 30ab :
Now consider the cube of a + 2b :
(a + 2b)3 = a3 + (2b)3 + 3× a×2b× (a + 2b) = a3 + 8b3 + 6ab×(a + 2b)
⇒ 53 = a3 + 8b3 + 6ab×5 [∵ a + 2b = 5 ]
⇒ 125 = a3 + 8b3 + 30ab
Thus the value of a3 + 8b3 + 30ab is 125.

6. If (a + 1/a)2 = 3 and a ≠ 0 then show that: a3 + 1/a3 = 0

7. If a + 2b + c = 0; then show that : a3 + 8b3 + c3 = 6abc.

Given that a + 2b + c = 0;
⇒ a + 2b = -c ….(1)
Now consider the expansion of ( a + 2b )3 :
(a + 2b)= (-c)3
⇒ a3 + (2b)3 + 3×a×2b×(a + 2b) = -c3

⇒ a3 + 8b3 + 3×a×2b×(-c) = -c3 [from (1)]

⇒ a3 + 8b3 – 6abc = -c

⇒ a3 + 8b3 – c3 = 6abc

Hence, proved.

8.1. Use property to evaluate : 133 + (-8)3 + (-5)3

Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 13, b = -8 and c = -5
⇒ 133 + (-8)3 + (-5)= 3(13)(-8)(-5) = 1560

8.2. Use property to evaluate : 73 + 33 + (-10)3

Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 7, b = 3, c = -10
⇒ 73 + 33 + (-10)= 3(7)(3)(-10) = -630

8.3. Use property to evaluate : 93 – 53 – 43

Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 9, b = -5, c = -4
⇒ 93 – 53 – 4= 93 + (-5)3 + (-4)= 3(9)(-5)(-4) = 540

8.4. Use property to evaluate : 383 + (-26)3 + (-12)3

Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc
a = 38, b = -26, c = -12
⇒ 383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568

9.1. If a ≠ 0 and a – 1/a = 3; find :

(i) a2 + 1/a2

9.2. If a ≠ 0 and a – 1/a = 3; Find : (ii) a3 − 1/a3

10.1. If a ≠ 0 and a – 1/a = 4 ; find : a2 + 1/a2

10.2. If a ≠ 0 and a – 1/a = 4 ; find :  a4 + 1/a4

10.3. If a ≠ 0 and a – 1/a = 4 ; find : a3 – 1/a3

11. If x ≠ 0 and x – 1/x = 2 ; then show that : x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4

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

Given that 2x – 3y = 10, xy = 16

∴ (2x – 3y)3 = (10)3

⇒ 8x3 – 27y3 – 3 (2x) (3y) (2x – 3y) = 1000

⇒ 8x3 – 27 y3 -18xy (2x – 3y) = 1000

⇒ 8x3 – 27 y3 – 18 (16) (10) = 1000

⇒ 8x3 – 27 y3 – 2880 = 1000

⇒8x3 – 27 y3 = 1000 + 2880

⇒ 8x3 – 27 y3 =3880

13.1. Expand : (3x + 5y + 2z) (3x – 5y + 2z)

(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

13.2. Expand : (3x – 5y – 2z) (3x – 5y + 2z)

(3x – 5y – 2z) (3x – 5y + 2z)

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

= (3x – 5y)2 – (2z)[since, (a + b) (a – b) = a2 – b2]

= 9x2 + 25y2 – 2 × 3x × 5y – 4z2

= 9x2 + 25y2– 30xy – 4z2

= 9x2 +25y2 – 4z2 – 30xy

14. The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes

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)= 92
⇒ a+ b+ 2ab = 81
⇒ a+ b+ 40 = 81
So, sum of squares is 81 – 40 = 41

Cubing on both sides gives
(a + b)= 93
⇒ a+ b+ 3ab(a + b) = 729
⇒ a+ b+ 60(9) = 729
⇒ a+ b= 729 – 540 = 189
So, the sum of cubes is 189.

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.

Given x – y = 5 and xy = 24 (x>y)
(x + y)= (x – y)+ 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.

Cubing on both sides gives
(x – y)= 53
⇒ x– y– 3xy(x – y) = 125
⇒ x– y– 72(5) = 125
⇒ x– y3= 125 + 360 = 485
So, difference of their cubes is 485.

Cubing both sides, we get
(x + y)= 113
⇒ x+ y+ 3xy(x + y) = 1331
⇒ x+ y= 1331 – 72(11) = 1331 – 792 = 539
So, sum of their cubes is 539.

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

xy = ab ….(i)

⇒ 4x+ y= a ….(ii)

Now, (2x + y)2 = (2x)2 + 4xy + y2

= 4x2 + y2 + 4xy

= a + 4b  [From (i) and (ii)]

⇒ 2x + y = ±√a+4b

### Exercise 4(C)

1.1. Expand : (x + 8) (x + 10)

(x + 8)(x + 10) = x2 + (8 + 10)x + 8×10
= x2 + 18x + 80

1.2. Expand : (x + 8)(x – 10)

(x + 8)( x – 10) = x2 + (8 – 10)x + 8×(-10)
= x2 – 2x – 80

1.3. Expand: (x – 8) (x + 10)

(x – 8) (x + 10) = x2 – (8 – 10)x – 8×10
= x2 + 2x – 80

1.4. Expand : (x – 8)(x – 10)

(x – 8)(x – 10) = x2 – (8 + 10)x + 8×10
= x2 – 18x + 80

2.1. Expand : (2x – 1/x)(3x + 2/x)

2.2. Expand : (3a + 2b)(2a – 3b)

3.1. Expand : (x + y – z)2

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

3.2. Expand : (x – 2y + 2)

(x – 2y + 2)2 = x2 + (2y)2 + (2)2 – 2(x)(2y) – 2(2y)(2) + 2(2)(x)
= x2 + 4y2 + 4 – 4xy – 8y + 4x

3.3. Expand : (5a – 3b + c)2

(5a – 3b + c)2 = (5a)2 + (3b)2 + (c)2 – 2(5a)(3b) – 2(3b)(c) + 2(c)(5a)

= 25a2 + 9b2 + c2 – 30ab – 6bc + 10ca

3.4. Expand : (5x – 3y – 2)2

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

= 25x2 + 9y2 + 4 – 30xy + 12y – 20x

3.5. Expand : (x – 1/x + 5)

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

We know that
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) ….(1)
Given that, a2 + b2 + c2 = 50 and a + b + c = 12.
We need to find ab + bc + ca:
Substitute the values of (a2 + b2 + c2 ) and (a + b + c)
in the identity (1), we have
(12)2 = 50 + 2(ab + bc + ca)

⇒ 144 = 50 + 2(ab + bc + ca)
⇒ 94 = 2( ab + bc + ca)
⇒ ab + bc + ca = 94/2
⇒ ab + bc + ca = 47

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

We know that
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) ….(1)
Given that, a2 + b2 + c2 = 35 and ab + bc + ca = 23
We need to find a + b + c:
Substitute the values of (a2 + b2 + c2) and (ab + bc + ca)
in the identity (1), we have
(a + b + c)2 = 35 + 2(23)
⇒ (a + b + c)2 = 81
⇒ a + b + c = ±√81
⇒ a + b + c = ±9

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

We know that
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) …..(1)
Given that, a + b + c = p and ab + bc + ca = q
We need to find a2 + b2 + c2 :
Substitute the values of (ab + bc + ca) and (a + b + c)
in the identity (1), we have
(p)2 = a2 + b2 + c2 + 2q
⇒ p2 = a2 + b2 + c2 + 2q
⇒ a2 + b2 + c2 = p2 – 2q

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

a2 + b2 + c2 = 50 and ab + bc + ca = 47
Since (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
∴ (a + b +  )2 = 50 + 2(47)
⇒ (a + b + c)2 = 50 + 94 = 144
⇒ a + b +c = √144
∴ a + b + c = ±12

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

x + y – z = 4 and x2 + y2 + z2 = 30
Since (x + y – z)2 =  x2 + y2 + z2 + 2(xy – yz – zx), we have
(4)2 = 30 + 2(xy – yz – zx)
⇒ 16 = 30 + 2( xy – yz – zx )
⇒ 2(xy – yz – zx) = -14
⇒ xy – yz – zx = −14/2= -7
∴ xy – yz – zx = -7

### Exercise 4(D)

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

[(x + 2y)2/xy + (2y + 3z)2/yz + (3z +x)2/zx]

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

2.1. If a + 1/a = m and a ≠ 0 ; find in terms of ‘m’; the value of : a- 1/a

2.2. If a + 1/a= m and a ≠ 0 ; find in terms of ‘m’; the value of : a2 - 1/a2

3.1. In the expansion of (2x2 – 8) (x – 4)2; find the value of coefficient of x3

(2x2 – 8)(x – 4)2
= (2x2 – 8)(x2 – 8x + 16)
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence, Coefficient of x3 = -16

3.2. In the expansion of (2x2 – 8) (x – 4)2; find the value of coefficient of x2

(2x2 – 8)(x – 4)2
= (2x2 – 8)(x2 – 8x + 16)
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence, Coefficient of x2 = 24

3.3. In the expansion of (2x2 – 8) (x – 4)2; find the value of constant term.

(2x2 – 8)(x – 4)2
= (2x2 – 8)(x2 – 8x + 16)
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence, Constant term = -128

4. If x > 0 and x2 + 1/9x2 = 25/36, Find : x3 + 1/27x3

5. If 2( x2 + 1 ) = 5x, find : (i) x – 1/x (ii) x3 – 1/x

6.1. If a2 + b2 = 34 and ab = 12; find : 3(a + b)2 + 5(a – b)2

a2 + b2 = 34, ab= 12

(a + b)2 = a2 + b2 + 2ab
= 34 + 2×12 = 34 + 24 = 58

(a – b)2 = a2 + b2 – 2ab
= 34 – 2×12 = 34- 24 = 10

3(a + b)2 + 5(a – b)2 = 3×58 + 5×10 = 174 + 50 = 224

6.2. If a2 + b2 = 34 and ab = 12; find : 7(a – b)2 – 2(a + b)2

a2 + b2 = 34, ab= 12

(a + b)2 = a2 + b2 + 2a
= 34 + 2×12 = 34 + 24 = 58

(a – b)2 = a2 + b2 – 2ab
= 34 – 2×12 = 34- 24 = 10\

7(a – b)2 – 2(a + b)2 = 7×10 – 2×58 = 70 – 116 = – 46

7. If 3x –4/x = 4 and x ≠ 0 ; find : 27x3 – 64/x3

8. If  x2 + 1/x2 = 7 and x ≠ 0;  find the value of  7x3 + 8x – 7/x3 – 8/x

9. If x = 1/(x −5) ; and x ≠ 5, find : x2 – 1/x2

10. If x = 1/(5−x) ; and x ≠ 5, find : x3 – 1/x3

11. If 3a + 5b + 4c = 0, show that : 27a3 + 125b3 + 64c3 = 180 abc

Given that 3a + 5b + 4c = 0
3a + 5b = – 4c
Cubing both sides,
(3a + 5b)3 = (-4c)3
⇒ (3a)3 + (5b)3 + 3×3a×5b (3a + 5b) = -64c3
⇒ 27a3 + 125b3 + 45ab×(-4c) = -64c3
⇒ 27a3 + 125b3 – 180abc = -64c3
⇒ 27a3 + 125b3 + 64c3 = 180abc
Hence proved.

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

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= 10

Now a2 + b2 = (a + b)2 – 2ab
= 72 – 2×10 = 49 – 20 = 29

13.1. Find the value of ‘a’: 4x2 + ax + 9 = (2x + 3)2

4x2 + ax + 9 = (2x + 3)2
Comparing coefficients of x terms, we get
ax = 12x
so, a = 12

13.2. Find the value of ‘a’: 4x2 + ax + 9 = (2x – 3)2

4x2 + ax + 9 = (2x – 3)2
Comparing coefficients of x terms, we get
ax = -12x
so, a = -12

13.3. Find the value of ‘a’: 9x2 + (7a – 5)x + 25 = (3x + 5)2

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

14.1. If (x2 +1)/x = 3(1/3) and x > 1; find x – 1/x

14.2. If (x2 +1)/x = 3(1/3) and x > 1; Find  x3 – 1/x3

15.1. The difference between two positive numbers is 4 and the difference between their cubes is 316. Find : Their product

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
a– b= 316
Cubing both sides,
(a – b)= 64
⇒ a– b– 3ab(a – b) = 64

Given a– b= 316
So, 316 – 64 = 3ab(4)
⇒ 252 = 12ab
So ab = 21; product of numbers is 21

15.2. The difference between two positive numbers is 4 and the difference between their cubes is 316. Find : The sum of their squares

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 …..(1)
a– b= 316 …..(2)

Squaring(eq 1) both sides, we get
(a – b)= 16
⇒ a+ b– 2ab = 16
⇒ a+ b= 16 + 42 = 58
Sum of their squares is 58.

### Exercise 4(E)

1.1. Simplify: (x + 6)(x + 4)(x – 2)

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

(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

1.2. Simplify : (x – 6)(x – 4)(x + 2)

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

(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

1.3. Simplify : ( x – 6 )( x – 4 )( x – 2 )

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

(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

1.4. Simplify : (x + 6)(x – 4)(x – 2)

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

(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

2.1. Simplify using following identity : (a ± b)(a2 ± ab + b2 ) = a³ ± b³
(x + 3y)(4x2 - 6xy + 9y2)

(2x+ 3y)(4x2 - 6xy + 9y2)
= (2x + 3y)[(2x)2 – (2x)(3y) + (3y)2]
= (2x)3 + (3y)3
= 8x3 + 27y3

2.2. Simplify using following identity : (a ± b)(a2 ± ab + b2) = a³ ± b³
(3x – 5/x)(9x2 + 15 + 25/x2 )

2.3. Simplify using following identity : (a ± b)(a2 ± ab + b2) = a³ ± b³
(a/3 −3b)(a2/9 + ab + 9b2

3.1. Using suitable identity, evaluate (104)3

Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(104)3 = (100 + 4)3
= (100)3 + (4)3 + 3×100×4(100 + 4)
= 1000000 + 64 + 1200×104
= 1000000 + 64 + 124800
= 1124864

3.2. Using suitable identity, evaluate (97)3

(97)= (100 – 3)3
= (100)3 – (3)3 – 3×100×3(100 – 3)
= 1000000 – 27 – 900 × 97
= 1000000 – 27 – 87300
= 912673

4. Simplify :(x2 – y2)3 + (y2 –z2)3 + (z2 –x2 )3/(x−y)3 + (y – z)3 + (z – x)3

5.1. Evaluate :(0.8×0.8×0.8 + 0.5×0.5×0.5)/(0.8×0.8 – 0.8×0.5 + 0.5×0.5)

5.2. Evaluate : (1.2×1.2 + 1.2×0.3 + 0.3×0.3)/(1.2×1.2×1.2 – 0.3×0.3×0.3)

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

a– 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
Since a – 2b + 3c = 0, we have
a– 8b3 + 27c= a3 + (-2b)3 + (3c)3
= 3(a)(-2b)(3c)
= -18abc

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

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

8. If x = 3 + 2√2, find : (i) 1/x

(ii) x – 1/x

(iii) (x – 1/x)3

(iv)x3 – 1/x3

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

a + b = 11 and a2 + b2 = 65
Now, (a+b)2 = a2 + b2 + 2ab
⇒ (11)2 = 65 + 2ab
⇒ 121 = 65 + 2ab
⇒ 2ab = 56
⇒ ab = 28

a3 + b3 = (a + b)(a2 – ab + b2)
= (11)(65 – 28)
= 11×37
= 407

10. Prove that : x2+ y2 + z2 – xy – yz – zx is always positive.

x+ y+ z– xy – yz – zx

= 2(x+ y+ z– xy – yz – zx)

= 2x+ 2y+ 2z– 2xy – 2yz – 2zx

= x+ x2 + y+ y2 + z2 + z– 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.

11.1. Find : (a + b)(a + b)

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

11.2. Find : (a + b)(a + b)(a + b)

(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

11.3. Find : (a – b)(a – b)(a – b)