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
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.