NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots.

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 7
Chapter Name	Cubes and Cube Roots
Exercise	Ex 7.1, Ex 7.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

Chapter 7 Cubes and Cube Roots Exercise 7.1

Question 1.
Which of the following numbers are not perfect cubes ?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Solution:
(i) Resolving 216 into prime factors, we find that 2
16 = 2x2x2 x 3x3x3
Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over.
∴ 216 is a perfect cube.

(ii) Resolving 128 into prime factors, we find that
128 = 2x2x2 x 2x2x2 x 2
Now, if we try to group together triples of equal factors, we are left with a single factor, 2.
∴ 128 is not a perfect cube.

(iii) Resolving 1000 into prime factors, we find that
1000 = 10 x 10 x 10 = 2x5x2x5x2x5 = 2x2x2x 5x5x5
Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over.
∴ 1000 is a perfect cube.

(iv) Resolving 100 into prime factors, we find that
100 = 10×10=2x5x2x5=2x2x5x5
Clearly, the prime factors of 100 cannot be grouped into triples of equal
factors.
∴ 100 is not a perfect cube.

(v) Resolving 46656 into prime factors, we find that
46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x3 x 3x3x3
Clearly, the prime factors of 46656 can be grouped into prime factors and no factor is left over.
∴ 46656 is a perfect cube.

Question 2.
Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Solution:
(i) Writing 243 as a product of prime factors, we have
243 = 3 x 3 x 3 x 3 x 3
Clearly, to make it a perfect cube, it must be multiplied by 3.

(ii) Writing 256 as a product of prime factors, we have
256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Clearly, to make it a perfect cube, it must be multiplied by 2.

(iii) Writing 72 as a product of prime factors, we have
72 = 2x2x2 x 3 x 3
Clearly, to make it a perfect cube, it must be multiplied by 3.

(iv) Writing 675 as a product of prime factors, we have
675 = 3 x 3 x 3 x 5 x 5
Clearly, to make it a perfect cube, it must be multiplied by 5.

(v) Writing 100 as a product of prime factors, we have
100 = 2 x 2 x 5 x 5
Clearly, to make it a perfect cube, it must be multiplied 2 x 5, i.e., 10.

Question 3.
Find the smallest number by which each of the following numbers must he divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Solution:
(i) Writing 81 as a product of prime factors, we have
81 = 3 x 3 x 3 x 3
Clearly, to make it a perfect cube, it must be divided by 3.

(ii) Writing 128 as a product of prime factors, we have
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Clearly, to make it a perfect cube, it must be divided by 2.

(iii) Writing 135 as a product of prime factors, we have
135 = 3 x 3 x 3 x 5
Clearly, to make it a perfect cube, it must be divided by 5.

(iv) Writing 192 as a product of prime factors, we have
192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Clearly, to make it a perfect cube, it must be divided by 3.

(v) Writing 704 as a product of prime factors, we have
704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Clearly, to make it a perfect cube, it must be divided by 11.

Question 4.
Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution:
Volume of the cuboid = 5 x 2 x 5 = 2 x 5 x 5
To make it a cube, we require 2 x 2 x 5, i.e., 20 such cuboids.

Chapter 7 Cubes and Cube Roots Exercise 7.2

Question 1.
Find the cube root of each of the following numbers by prime factorisation method :
(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125
Solution:
(i) Resolving 64 into prime factors, we get
64 = 2 x 2 x 2 x 2 x 2 x 2
∴

(ii) Resolving 512 into prime factors, we get
512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
∴

(iii) Resolving 10648 into prime factors, we get
10648 = 2 x 2 x 2 x 11 x 11 x 11
∴

(iv) Resolving 27000 into prime factors, we get
27000 = 1000 x 27
= 10 x 10 x 10 x 3 x 3 x 3 = 2 x 5 x 2 x 5 x 2 x 5 x 3 x 3 x 3
= 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5
∴

(v) Resolving 15625 into prime factors, we get
15625 = 5 x 5 x 5 x 5 x 5 x5
∴

(vi) Resolving 13824 into prime factors, we get
13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
∴

(vii) Resolving 110592 in prime factors, we get
110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
∴

(viii) Resolving 46656 in prime factors, we get
46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
∴

(ix) Resolving 175616 in prime factors, we get
175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7 = (2 x 2 x 2 x 7)
= 56

(x) Resolving 91125 in prime factors, we get
91125 = 3 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5
∴

Question 2.
State true or false :
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Solution:
(i) False
(ii) True
(iii) False, as 15² = 225 and 15³ = 3375
(iv) False, as 8 = 2³, 1728 =12³, etc.
(v) False
(vi) False, as 10³ = 1000, 99³ = 970299
(vii) True, as 1³ = 1, 2³ =8.

Question 3.
You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
For 1331 :
Units digit of the cube root of 1331 is 1 as unit’s digit of the cube root of numbers ending in 1 is 1 . After striking three digits from the right of 1331, we get the number 1. Since 1³ = 1, so the ten’s digit of the cube root of given number is 1.
∴

For 4913 :
Units digit of the cube root of 4913 is 7 as unit’s digit of cube root of numbers ending in 3 is 7. After striking three digits from the right of 4913, we get the number 4. As 1³ =1 and 2³ =8, so 1³ < 4 < 2³. Therefore, the ten’s digit of cube root of 4913 is 1.
∴

For 12167 :
Unit’s digit of the cube root of 12167 is 3 as unit’s digit of cube root of numbers ending in 7 is 3. After striking three digits from the right of 12167, we get the number 12. As 2³ =8 and 3³ =27, so 2³ < 12 < 3³. So, the ten’s digit of the cube root of 12167 is 2.
∴

For 32768 :
Unit’s digit of the cube root of 32768 is 2 as unit’s digit of cube root of numbers ending in 8 is 2. After striking three digits from the right of 32768, we get the number 32. As 3³ =27 and 4³ = 64, so 3³ < 32< 4³. So, the ten’s digit of the cube root of 32768 is 3.
∴

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