# Expansions ICSE Class-9th Concise Selina Mathematics

Expansions ICSE Class-9th Concise Selina Mathematics Solutions (Including Substitution) Chapter-4. We provide step by step Solutions of Exercise / lesson-4 Expansions (Including Substitution) for ICSE Class-9 Concise Selina Mathematics by RK Bansal.

Our Solutions contain all type Questions with Exe-4 A, Exe-4 B, Exe-4 C, Exe-4 D and Exe-4 E to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

## Expansions ICSE Class-9th Concise Selina Mathematics Solutions (Including Substitution) Chapter-4

–: Select Topics :–

Exe-4 A,

Exe-4 B,

Exe-4 C,

Exe-4 D,

### Exercise – 4(A)Expansions (Including Substitution) ICSE Class-9th Concise Selina Mathematics

#### Question 1.1

Find the square of : 2a + b

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

#### Question 1.2

Find the square of : 3a + 7b

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

#### Question 1.3

Find the square of : 3a – 4b

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

#### Question 1.4

Find the square of : ……….

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

#### Question 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 x 100 x 1
= 10,000 + 1 + 200
= 10,201

#### Question 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 x 500 x 2
= 250000 + 4 + 2000
= 2,52,004

#### Question 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 x 100 x 3
= 10000 + 9 – 600
= 9,409

#### Question 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 x 1000 x 2
= 1000000 + 4 – 4000
= 9,96,004

Evalute : …………

Evalute :………….

#### Question 4.1

Evaluate :…………..

Consider the given expression :

#### Question 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 x 4a x 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 x 4a x 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

#### Question 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

#### Question 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

#### Question 7

If x + y =    ; 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)

#### Question 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

#### Question 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

#### Question 10

If a + = 6 and  a ≠ 0 find :

(i) a….

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

#### Question 11

If a………….find :

(i) a….

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

#### Question 12

If a2 – 3a + 1 = 0, and a ≠ 0; find :

(i)……

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

#### Question 13

If a2 – 5a – 1 = 0 and a ≠ 0 ; find :
(i) ………
(ii) …………

(iii) ………..

(i) Consider the given equation
a2 – 5a – 1 = 0
Rewrite the given equation, we have
a2 – 1 = 5a

#### Question 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

#### 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.

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.

#### Question 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.

### Expansions (Including Substitution) Exe-4 B for ICSE Class-9th Concise Selina Mathematics

#### Question 1.1

Find the cube of : 3a- 2b

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

#### Question 1.2

Find the cube of : 5a + 3b

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

#### Question 1.3

Find the cube of :

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

#### Question 1.4

Find the cube of :

( a – b )3 = a3 – 3ab( a – b ) – b3

#### Question 2

If  a2 + ………….   find :
(i) ………
(ii) ……..

#### Question 3

If  a2 + ………….   find :
(i) ………
(ii) ……..

#### Question 4

If  ; then show that

…………

#### Question 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 x a x 2b x ( a + 2b )
= a3 + 8b3 + 6ab x ( a + 2b )
53 = a3 + 8b3 + 6ab x 5       [ ∵ a + 2b = 5 ]
125 = a3 + 8b3 + 30ab
Thus the value of a3 + 8b3 + 30ab is 125.

If =0

#### Question 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 x a x 2b x ( a + 2b ) = -c3

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

⇒                           a3 + 8b3 – 6abc = -c

⇒                              a3 + 8b3 – c3 = 6abc

Hence proved.

#### Question 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

#### Question 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

#### Question 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

#### Question 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

#### Question 9.1

If a ≠ 0 and = 3 ; find :

…………

#### Question 9.2

If a ≠ 0 and …….. 3 ; Find :

…………

#### Question 10.1

If a ≠ 0 and = 4 ; find : ………

#### Question 10.2

If a ≠ 0 and = 4 ; find : ………

#### Question 10.3

If a ≠ 0 and = 4 ; find : ………

#### Question 11

If X ≠ 0 and X + ………; then show that :

……….

#### Question 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

#### Question 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

#### Question 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)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

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.

#### 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.

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.

#### Question 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 =

### Selina Solutions of Expansions (Including Substitution) Exe-4 C for ICSE Class-9th Concise Mathematics

#### Question 1.1

Expand : ( x + 8 ) ( x + 10 )

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

#### Question 1.2

Expand : ( x + 8 )( x – 10 )

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

#### Question 1.3

Expand : ( X – 8 ) ( X + 10 )

( X – 8 ) ( X + 10 ) = X2 – ( 8 – 10 )X – 8 x 10
= X2 + 2X – 80

#### Question 1.4

Expand : ( X – 8 )( X – 10 )

( X – 8 )( X – 10 ) = X2 – ( 8 + 10 )X + 8 x 10
= X2 – 18X + 80

Expand :

Expand :

#### Question 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

#### Question 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

#### Question 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

#### Question 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

Expand :

#### Question 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 =
⇒ ab + bc + ca = 47

#### Question 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 =
⇒ a + b + c =

#### Question 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

#### Question 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 + c )2 = 50 + 2(47)
⇒ ( a + b + c )2 = 50 + 94 = 144
⇒ a + b +c =
∴ a + b + c =

#### Question 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 =
∴ xy – yz – zx = -7

### Concise Selina Maths Solutions, Exe-4 D Expansions (Including Substitution) ICSE Class-9th

#### Question 1

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

………………………………

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

#### Question 2.1

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

#### Question 2.2

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

………………….

#### Question 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

#### Question 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

#### Question 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

If x > 0 and

#### Question 5

If 2( x2 + 1 ) = 5x, find :
(i) ………

(ii) ……………

#### Question 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 x 12 = 34 + 24 = 58

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

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

#### Question 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 + 2ab
= 34 + 2 x 12 = 34 + 24 = 58

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

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

#### Question 7

f 3x – …………………………

#### Question 8

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

……………..

#### Question 9

If x =…………….. Find:…………..

#### Question 10

If x = ……… find………….

#### Question 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 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.

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 x 10 = 49 – 20 = 29

#### Question 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

#### Question 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

#### Question 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

If

If

#### Question 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

#### Question 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)Expansions (Including Substitution) ICSE Class-9th Concise Selina Mathematics

#### Question 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

#### Question 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

#### Question 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

#### Question 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

#### Question 2.1

Simplify using following identity : ( 4x2 + 6xy + 9y)

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

#### Question 2.2

Simplify using following identity :

#### Question 2.3

Simplify using following identity :

#### Question 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

#### Question 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

Simplify :………..

Evaluate :……………

Evaluate :

#### Question 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

#### Question 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

#### Question 8

If x = 3 + 2√2, find :
(i) ……………..

(ii) …………..

(iii) …………

(iv)……

#### Question 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 x 37
= 407

#### Question 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.

#### Question 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

#### Question 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

#### Question 11.3

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

replacing b by -b, we get
= a3 + (-b)3 + 3a2(-b) + 3a(-b)2
= a3 – b3 – 3a2b + 3ab2

— End of Expansions ICSE Class-9th Concise Solutions :–

Thanks