Selina Concise Class 9 Maths Chapter 4 Expansions 4B Solutions

Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4B Solutions

 

Exercise 4B

 

(1) Find the cube of

(i) 3a – 7b

=> Solution:- We have to cube of (3a – 7b)

Now, we have to use formula

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

also, we have to use

(a- b)3 = a3 – 3ab (a – b) + b3

∴ a = 3a and b = 7b

∴ (3a – 2b)3

 = (3a)3 – 3 × (3a) × (2b) (3a – 2b) + (-2b)3

= 9a3 – 18ab (3a – 2b) + (-8b3)

(3a – 2b)3 = 9a3 – 54a2b + 36ab2 – 8b3

 

(ii) 5a + 3b

=> Solution:-

We have to cube of (5a + 3b)

Now, we have to use formula.

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

∴ a = 5a and b = 3b

∴ (5a + 3b)3 = (5a)3 + 3 × (5a) × (3b) (5a + 3b) + (3b)3

= 125a3 + 45ab (5a + 3b) + 27b3

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

 

(iii) 2a + 1/2a

=> Solution:-

We have to cube of (2a + ½ a)

Now, we have to use formula,

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

∴ a = 2a and b = ½ a

∴ (2a + 1/2a) = (2a)3 + 3 × (2a) × (1/2a) (2a + 1/2a) + (1/2a)3

= 8a3 + 3 (2a + 1/2a) + 1/8 a3

(2a + 1/2a) = 8a3 + 6a + 3/2 a + 1/8 a3

 

(iv) 3a – 1/a (a ≠ 0)

=> Solution:-

We have to cube of (3a – 1/a)

Now, we have to use formula,

(a – b)3 = a3 – 3ab (a –b) + b3

∴ a = 3a and b = 1/a

∴ (3a -1/a)3 = (3a)3 – 3 × (3a) × (1/a) (3a – 1/a) + (-1/a)3

(3a – 1/a)3 = 27a3 – 9 (3a – 1/a) – 1/a3

(3a -1/a)3 = 27a3 – 27a + 9/a – 1/a3

 

(2) If a2 + 1/a2 =  47 and a  ≠ 0 find

(i) a + 1/a

(ii) a3 + 1/a3

=> Solution:- given that, a2 + 1/a2 = 47 and a  ≠ 0

(i) To find => a + 1/a

we have to use formula,

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

∴ (a + 1/a)2 = a2 + 2 × a × 1/a + (1/a)2

(a + 1/a)2 = (a2 + 1/a2) + 2

(a + 1/a)2 = 47 + 2

(a + 1/a)2 = 49

Taking square root on both sides,

√(a+1/a)2 = √49

(a + 1/a) = ± 7 …..(A)

 

(ii) To find = a3 + 1/a3

Now, we have to use formula-

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

∴ (a + 1/a)3 = a3 + 3 × a × 1/a (a + 1/a) + (1/a)3

= a3 + 3(a + 1/a) + 1/a3

= a3 + 1/a3 + 3 (± 7) (∵ from A)

(7)3 = a3 + 1/a3 ± 21

343 = a3 + 1/a3 ± 21

(± 343) – (±21) = a3 + 1/a3

± 322 = a3 + 1/a3

∴ a3 + 1/a3 = ± 322

 

(3) If a2 + 1/a2 = 18, a ≠ 0 find:

(i) a – 1/a

(ii) a3 – 1/a3

=> Solution:- given that, a2 + 1/a2 = 18 , a ≠ 0

(i) To find => a – 1/a

we have to formula,

(a –b)2 = a2 – 2ab + b2

∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2

= (a2 + 1/a2) – 2

(a – 1/a)2 = 18 -2 (∵ from given)

= 16

Taking square root on both sides,

√ (a -1/a)2 = √16

(a – 1/a) = ± 4 …….(A)

(ii) To find => a3 – 1/a3

Now, we have to use formula,

(a – b)3 = a3 – 3ab (a – b) + b3

∴ (a – 1/a)3 = a3 – 3 × a × 1/a (a – 1/a) + (- 1/a)3

(a – 1/a)3 = a3 – 3 (a -1/a) – 1/a3

(a – 1/a)3 = (a3 – 1/a3) -3 (a – 1/a)

(+4)3 = (a3 – 1/a3) – 3 (± 4) (∵ from A)

± 64 = (a3 – 1/a3) – (± 12)

(± 64) + (± 12) = a3 – 1/a3

± 76 = a3 – 1/a3

∴ a3 – 1/a3 = ± 76

 

(4) If a + 1/a = P and a ≠  0; then show that:

a3 + 1/a3 = P (P2 – 3)

=> Solution:- given that, a + 1/a = P and a ≠ 0

∴ a + 1/a = P

cubing on both side,

(a + 1/a)3 = P3

∴ a3 + 3a × 1/a (a + 1/a) + 1/a3 = P3 { ∵(a+ b)3 = a3 + 3ab (a + b) + b3}

a3 + 3(a + 1/a) 1/a3 = P3

(a3 + 1/a3) + 3 (a + 1/a) = P3

(a3 + 1/a3) + 3(P) = P3

(a3 + 1/a3) + 3P = P3 { ∵ from given}

(a3 + 1/a3) = P3 – 3P

(a3 + 1/a3) =P (P2 -3) Hence proved

 

(5) If  a + 2b = 5 , then show that

a3 + 8b3 + 30ab = 125

=> Solution:- given that, a + 2b = 5

cubing on both sides,

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

(a3 + 3a × (2b) (a + 2b) + (2b)3) = 125

a3 + 6ab (a + 2b) + 8b3 = 125

a3 + 6ab(5) + 8b3 = 125 (∵ from given)

a3 + 30ab + 8b3 = 125

∴ a3 + 8b3 + 30ab = 125 Hence Proved

 

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

a3 + 1/a3 = 0

=> Solution:- given that, (a + 1/a)2 = 3 and a ≠ 0

Taking square root on both sides-

√(a+ 1/a)2 = √3

a + 1/a = ± √3

Now, we have to use formula,

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

(a + 1/a)3 = a3 + 3a × 1/a (a + 1/a) + (1/a)3

(a + 1/a)3 = a3 + 3(a + 1/a) + 1/a3

(a + 1/a)3 = (a3 + 1/a3) + 3(a+ 1/a)

(a + 1/a)3 = (a3 + 1/a3) + 3 (a + 1/a)

(a + 1/a)3 = 3 (a + 1/a) + (a3 + 1/a3)

(±√3)3 = 3 (±√3) + (a3 + 1/a3)

(±√3) × (±√3) × (±√3) = 3 (±√3) + a3 + 1/a3

±3√3 = ±3√3 + a3 + 1/a3

(± 3√3) – (±3√3) = a3 + 1/a3

∴ 0 = a3 + 1/a3

∴ a3 + 1/a3 = 0 Hence Proved

 

(7) If a + 2b + c = 0 , then show that

a3 + 8b3 + c3 = 6abc

=> Solution:- given that, a + 2b + c = 0

Now, arrange given equation,

a + 2b = -c

Now, cubing on both sides,

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

a3 + 3a × (2b) (a + 2b) + (2b)3 = -c3 { ∵ (a + b)3 = a3 + 3ab (a +- b) + b3}

a3 + 6ab (a + 2b) + 8b3 = – c3

a3 + 6ab (-c) + 8b3 = -c3

a3 + 8b3 + c3 = 6abc   Hence Proved

 

(8) Use property to evaluate:-

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

=> Solution:-

Let a = 13

b = -8

c = -5

Now, a + b + c = 13 + (-8) + (-5)

= 13 – 8 – 5

13 – 13

a + b + c = 0

we know that,

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

=> (13)3 + (-8)3 + (-5)3 = 3 × (13) × (-8) × (-5)

= 39 × 40

= 1560

 

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

=> Solution:-

Let a = 7

b = 3

c = -10

Now, a + b + c = 7 + 3 + (-10)

= 10 – 10

a + b + c = 0

we know that,

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

=> (7)3 + (3)3 + (-10)3

= 3 × (7) × (3) × (-10)

= (21) × (-30)

= – 630

 

(iii) 93 – 53  – 43

=> Solution:- Let a = 9

b = – 5

c = – 4

Now, a + b + c = 9 + (-5) + (-4)

= 9 – 5 – 4

= 9 – 9

= 0

we know that,

if a + b + c = 0 then a3 + b3 + c3 = 3bca

∴ 93 – 53 – 43 = 3 × (-5) × (-4) × (9)

= (27) × (20)

= 540

 

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

=> Solution:- Let a = 38, b = -26, c = -12

Now, know that,

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

∴ (38)3 + (-26)3 + (-12)3 = 3 × (38) × (-26) × (-12)

= (114) × (312)

= 35,568

 

(9) If a ≠ 0 and a – 1/a = 3 ; find

(i) a2 + 1/a2

(ii) a3 – 1/a3

=> Solution :- given that, a ≠ 0 and a – 1/a = 3

Now, To find => (i) a2+ 1/a2

Now, we have to use,

(a – b)2 = a2 – 2ab + b2

∴ (a – 1/a)2 = a2 – 2 × a × 1/a + (1/a)2

(a – 1/a)2 = (a2 + 1/a2) – 2

(3)2 = (a2 + 1/a2) – 2 {∵ from given}

9 = (a2 + 1/a2) – 2

9 + 2 = a2 + 1/a2

11 = a2 + 1/a2

∴ a2 + 1/a2 = 11

 

(ii) To find => a3 – 1/a3

Now, we have to use,

(a –b)3 = a3 – 3ab (a – b) + b3

∴ (a -1/a)3 = a3 – 3 × a × 1/a (a – 1/a) + (-1/a)3

(a – 1/a)3 = a3 – 1/a3 – 3( a- 1/a)

(3)3 = (a3 – 1/a3) – 3 (3) {∵ from given}

27 = (a3 – 1/a3) – 9

27 + 9 = (a3 – 1/a3)

36 = a3 – 1/a3

∴ a3 – 1/a3 = 36

 

(10) If a ≠ 0 and a – 1/a = 4 ; find

(i) a2 + 1/a2

(ii) a4 + 1/a4

(iii) a3 – 1/a33

=> Solution:- given that, a -1/a = 4 and a ≠ 0

(i) To find => a2 + 1/a2

we have to use formula,

(a –b)2 = a2 – 2ab + b2

∴ (a – 1/a)2 = a2 – 2 × a × 1/a + ( 1/a)2

(a -1/a)2 = (a2 + 1/a2) -2

(4)2 = (a2 + 1/a2) -2 {∵ from given}

16 = a2 + 1/a2 – 2

16 + 2 = a2 + 1/a2

18 = a2 + 1/a2

∴ a2 + 1/a2 = 18……(A)

 

(ii) To find => a4 + 1/a4

first we have to rewrite a4 + 1/a4

(a2 + 1/a2)2

Now, we have to use formula,

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

from equation (A)

a2 + 1/a2 = 18

squaring on both sides-

(a2 + 1/a2)2 = (18)2

(a2)2 + 2 × a2 ×(1/a2) + (1/a2)2 = 324

a4 + 2 + 1/a4  = 324

a4 + 1/a4 = 324 – 2

a4 + 1/a4 = 322

(iii) To find => (iii) a3 – 1/a3

Now, we have to use formula,

(a- b)3 = a3 – 3ab (a –b) + b3

∴ (a -1/a)3 = a3 – 3 × a × 1/a (a – 1/a) + (-1/a)3

= a3 – 3 (a -1/a) – 1/a3

(a -1/a)3 = (a3 – 1/a3) – 3 (a -1/a)

(4)3 = (a3 – 1/a3) – 3(4)

64 = (a3 -1/a3) – 12

64 + 12 = a3 – 1/a3

76 = a3 – 1/a3

∴ a3 – 1/a3 = 76

 

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

=> Solution:- given that,

2x – 3y = 10 …..(1) and

xy = 16

To find-  8x3 – 27y3

So, cube on both side of equation (1),

we get,

2x – 3y = 10

(2x – 3y)3 = (10)3

by using the formula,

(a –b)3 = a3 – 3ab (a –b) + b3

∴ (2x)3 – 3(2x) × (3y) (2x – 3y) + (-3y)3 =10,00

8x3 – 18xy (2x – 3y) + (-27y3) = 10,00

8x3 – 18 × 16 (2x – 3y) – 27y3 = 10,00 (∵ given xy = 16)

8x3 – 18  × 16  × (10) – 27y3 = 1000 (∵ given 2x -3y = 10)

8x3 – 2880 – 27y3 = 1000

8x3 – 27y3 = 1000 + 2880

8x3 – 27y3 = 3880

 

(13) Expand

(i) (3x + 5y + 2z) (3x – 5y + 2z)

(ii) (3x – 5y -2z) (3x – 5y + 2z)

=> (i) (3x + 5y + 2z) (3x – 5y + 2z)

Multiply this two brackets

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

= 9x2 – 15xy + 6xz + 15xy – 25y2 + 10yz + 6xz – 10yz + 4z2

= 9x2 – 25y2 + 4z2 – 15xy + 6xz + 15xy + 10yz + 6xz – 10yz

= 9x2 – 25y2 + 4z2 + 12xz

(ii) (3x – 5y – 2z) (3x – 5y + 2z)

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

= 9x2 – 15xy + 6xz – 15xy + 25y2 – 10yz – 6xz + 10yz – 4z2

= 9x2 + 25y2 – 4z2 – 15xy + 6xz – 15xy – 10yz – 6xz + 10yz

= 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

=> Solution:- Let the two numbers are x and y

given that, The sum of two numbers is 9

∴ x + y = 9 …….(1)

and also given that, their product is 20

xy = 20 ……(2)

(i) To find the sum of squares.

So, squaring on both sides of equation (1)

(x + y)2 = (9)2

we have to use formula,

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

(x2 + 2xy + y2) = 81

x2 + 2xy + y2 = 81

x2 + 2 × 20 + y2 = 81 (∵ given xy = 20)

x2 + y2 + 40 = 81

x2 + y2 = 81 – 40

x2 + y2 = 41

(ii) To find the sum of their cubes-

So, cubing on both sides of equation(1)

(x + y)3 = (9)3

we have to use formula,

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

∴ (x3 + 3xy (x+ y) + y3) = 729

x3 + 3 × 20 × 9 = 729 (∵ given x + y = 9 xy = 20)

x3 + 540 = 729

x3 = 729 – 540

x3 = 189

∴ The sum of their 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

=> Solution:- given that, Two positive numbers x and y are such that x > y

first condition, If the difference of these numbers is 5

∴ x – y = 5 …. (1)

and second condition is , their product is 24

xy = 24 …..(2)

(i) To find sum of these numbers,

(x + y)2 = (x – y)2 + 4xy { If we have to solve this then the remaining term is only 4xy}

∴ (x + y)2 = (5)2 + 4 × 24 {∵ given xy = 24}

= 25 + 96

(x + y)2 = 121

Taking square on both sides,

√ (x + y)2 = √121

(x + y) = 11…..(A)

(ii) To find Difference of their cubes

cubing on both sides of equation (1),

we have to use the formula,

(a – b)3 = a3 – 3ab (a –b) + b3

∴ x3 – 3xy (x –y) – y3 = 125

x3 – 3 × 24 × 5 – y3 = 125 {∵ given xy = 24 (x – y) = 5}

x3 – 360 – y3 = 125

x3 – y3 = 125 + 360

x3 – y3 = 485

 

(iii) To find sum of their cubes,

cubing on both sides of equation (A),

(x + y)3 = (11)3

we have to use formula,

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

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

x3 + y3 + 3 × 24 × 11 = 1331

x3 + y3 + 792 = 1331

x3 + y3 = 1331 – 792

x3 + y3 = 539

 

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

=> Solution:- given that, 4x2 + y2 = a and xy = b

To find  2x + y

Squaring of (2x + y)2 = (2x)2 + 2 × (2x) y + y2 {∵ (a +b)2 = a2 + 2ab + b2}

= 4x2 + 4xy + y2

= 4x2 + 4 × b + y2

= 4x2 + y2 + 4b

(2x + y)2 = a + 4b { ∵ given 4x2 + y2 = a and xy = b}

Taking square root on both sides,

√ (2x+y)2 = √ a+ 4b

2x + y =  ± √a + 4b

 

 

Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4B

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