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

CBSE Solutions | Maths Class 8

Check the below NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Pdf free download. NCERT Solutions Class 8 Maths were prepared based on the latest exam pattern. We have Provided Cubes and Cube Roots Class 8 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 8 Maths

NCERT Solutions Class 8 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	8th
Subject:	Maths
Chapter:	7
Chapters Name:	Cubes and Cube Roots
Medium:	English

Cubes and Cube Roots | Class 8 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.1

Ex 7.1 Class 8 Maths 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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 1

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 2

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 3

Ex 7.1 Class 8 Maths 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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 4

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 5

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 6

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

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

Ex 7.1 Class 8 Maths 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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 9

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 10

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 11

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 12

(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.
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 13

Ex 7.1 Class 8 Maths 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.

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Exercise 7.2

Ex 7.2 Class 8 Maths 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
∴ \(\sqrt [ 3 ]{ 64 } =\left( 2\times 2 \right) =4\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 14

(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
∴ \(\sqrt [ 3 ]{ 512 } =\left( 2\times 2\times 2 \right) =8\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 15

NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 15

(iii) Resolving 10648 into prime factors, we get
10648 = 2 x 2 x 2 x 11 x 11 x 11
∴ \(\sqrt [ 3 ]{ 10648 } =\left( 2\times 11 \right) =22\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 16

(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
∴ \(\sqrt [ 3 ]{ 27000 } =2\times 3\times 5=30\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 17

(v) Resolving 15625 into prime factors, we get
15625 = 5 x 5 x 5 x 5 x 5 x5
∴ \(\sqrt [ 3 ]{ 15625 } =5\times 5=25\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 18

(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
∴ \(\sqrt [ 3 ]{ 13824 } =\left( 2\times 2\times 2\times 3 \right) =24\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 19

(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
∴ \(\sqrt [ 3 ]{ 110592 } =\left( 2\times 2\times 2\times 2\times 3 \right) =48\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 20

(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
∴ \(\sqrt [ 3 ]{ 46656 } =\left( 2\times 2\times 3\times 3 \right) =36\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 21

(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
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 22

(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
∴ \(\sqrt [ 3 ]{ 91125 } =\left( 3\times 3\times 5 \right) =45\)
NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots 23

Ex 7.2 Class 8 Maths 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.

Ex 7.2 Class 8 Maths 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.
∴ \(\sqrt [ 3 ]{ 1331 } =11\)

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.
∴ \(\sqrt [ 3 ]{ 4913 } =17\)

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.
∴ \(\sqrt [ 3 ]{ 12167 } =23\)

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.
∴ \(\sqrt [ 3 ]{ 32768 } =32\)