NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.1

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.1 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Chapter 6 teaches what a cube of a number is and how to find a cube root. When you multiply a number by itself three times, you get its cube. Exercise 6.1 is about cubes and perfect cubes. Exercise 6.2 goes on to cube roots.

This exercise has 4 questions. All of them are solved using one method, called prime factorisation. You break a number into its prime factors, then group those factors in threes. If every factor forms a complete group of three, the number is a perfect cube. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.1

Cubes and Cube Roots
Medium
Q.
Which of the following numbers are not perfect cubes?

(i) 216 (ii) 128 (iii) 1000 (iv)100 (v) 46656
Cubes and Cube Roots
Medium
Q.
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
Cubes and Cube Roots
Medium
Q.
Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704
Cubes and Cube Roots
Difficult
Q.
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?
Cubes and Cube Roots
Medium
Q.
Find the cube root of each of the following numbers by prime factorization method.

(i) 64 (ii) 512 (iii) 10648 (iv)27000 (v) 15625
(vi)13824 (vii) 110592 (viii) 46656 (ix) 175616
(x) 91125
Cubes and Cube Roots
Medium
Q.
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.
Cubes and Cube Roots
Difficult
Q.
You are told that 1,331 is a perfect cube. Can you guess without factorization what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.1

Exercise 6.1 – Questions and Answers

Q 1. Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656

Answer:

(i) The prime factorisation of 216 is as follows:
2 216
2 108
2 54
3 27
3 9
3 3
1

216 = (2 × 2 × 2) × (3 × 3 × 3)
Since all the factors appear in a group of three, so 216 is a perfect cube.

(ii) The prime factorisation of 128 is as follows:
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

128 = (2 × 2 × 2) × (2 × 2 × 2) × 2
Since 2 is not appearing in a group of three, so 128 is not a perfect cube.

(iii) The prime factorisation of 1000 is as follows:
2 1000
2 500
2 250
5 125
5 25
5 5
1

1000 = (2 × 2 × 2) × (5 × 5 × 5)
Since 2 and 5 appear in a group of three, so 1000 is a perfect cube.

(iv) The prime factorisation of 100 is as follows:
2 100
2 50
5 25
5 5
1

100 = 2 × 2 × 5 × 5
Since 2 and 5 are not appearing in a group of three, so 100 is not a perfect cube.

(v) The prime factorisation of 46656 is as follows:
2 46656
2 23328
2 11664
2 5832
2 2916
2 1458
3 729
3 243
3 81
3 27
3 9
3 3
1

46656 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)
Since all the factors appear in a group of three, so 46656 is a perfect cube.

Therefore, the numbers that are not perfect cubes are (ii) 128 and (iv) 100.

Q 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

Answer:

(i) The prime factorisation of 243 is as follows:
3 243
3 81
3 27
3 9
3 3
1

243 = (3 × 3 × 3) × 3 × 3
Here, two 3s are left which do not appear in a group of three. To make 243 a cube, one more 3 is required.
In that case, we have to multiply 243 by 3
i.e. 243 × 3 = (3 × 3 × 3) × (3 × 3 × 3) = 729, which is a perfect cube.
Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) The prime factorisation of 256 is as follows:
2 256
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

256 = (2 × 2 × 2) × (2 × 2 × 2) × 2 × 2
Here, two 2s are left which do not appear in a group of three. To make 256 a cube, one more 2 is required.
If we multiply 256 by 2, we get
256 × 2 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 512, which is a perfect cube.
Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) The prime factorisation of 72 is as follows:
2 72
2 36
2 18
3 9
3 3
1

72 = (2 × 2 × 2) × 3 × 3
Here, two 3s are left which are not in a group of three. To make 72 a perfect cube, one more 3 is required.
Then, we obtain
72 × 3 = (2 × 2 × 2) × (3 × 3 × 3) = 216, which is a perfect cube.
Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) The prime factorisation of 675 is as follows:
3 675
3 225
3 75
5 25
5 5
1

675 = (3 × 3 × 3) × 5 × 5
Here, two 5s are left which are not in a group of three. To make 675 a cube, one more 5 is required.
If we multiply 675 by 5, we get
675 × 5 = (3 × 3 × 3) × (5 × 5 × 5) = 3375, which is a perfect cube.
Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) The prime factorisation of 100 is as follows:
2 100
2 50
5 25
5 5
1

100 = 2 × 2 × 5 × 5
Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5.
Then, we obtain
100 × 2 × 5 = (2 × 2 × 2) × (5 × 5 × 5) = 1000, which is a perfect cube.
Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 × 5 = 10.

Q 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

Answer:

(i) The prime factorisation of 81 is as follows:
3 81
3 27
3 9
3 3
1

81 = (3 × 3 × 3) × 3
Here, one 3 is left which is not in a triplet.
If we divide 81 by 3, then it will become a perfect cube.
Therefore, 81 ÷ 3 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

(ii) The prime factorisation of 128 is as follows:
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

128 = (2 × 2 × 2) × (2 × 2 × 2) × 2
Here, one 2 is left which is not in a group of three.
If we divide 128 by 2, then it will become a perfect cube.
Therefore, 128 ÷ 2 = 64 = (2 × 2 × 2) × (2 × 2 × 2) is a perfect cube.
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

(iii) The prime factorisation of 135 is as follows:
3 135
3 45
3 15
5 5
1

135 = (3 × 3 × 3) × 5
Here, one 5 is left which is not in a group of three.
If we divide 135 by 5, then it will become a perfect cube.
Thus, 135 ÷ 5 = 27 = 3 × 3 × 3 is a perfect cube.
Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.

(iv) The prime factorisation of 192 is as follows:
2 192
2 96
2 48
2 24
2 12
2 6
3 3
1

192 = (2 × 2 × 2) × (2 × 2 × 2) × 3
Here, one 3 is left which is not in a group of three.
If we divide 192 by 3, then it will become a perfect cube.
Thus, 192 ÷ 3 = 64 = (2 × 2 × 2) × (2 × 2 × 2) is a perfect cube.
Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.

(v) The prime factorisation of 704 is as follows:
2 704
2 352
2 176
2 88
2 44
2 22
11 11
1

704 = (2 × 2 × 2) × (2 × 2 × 2) × 11
Here, one 11 is left which is not in a group of three.
If we divide 704 by 11, then it will become a perfect cube.
Thus, 704 ÷ 11 = 64 = (2 × 2 × 2) × (2 × 2 × 2) is a perfect cube.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.

Q 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?

Answer:

Volume of the cuboid of sides 5 cm, 2 cm, 5 cm
= 5 cm × 2 cm × 5 cm = 5 × 5 × 2 cm³

Here, two 5s and one 2 are left which are not in a triplet. If we multiply this expression by 5 × 2 × 2 = 20, then it will become a perfect cube.

Thus, 5 × 5 × 2 × 5 × 2 × 2 = (2 × 2 × 2) × (5 × 5 × 5) = 1000, which is a perfect cube.

Hence, 20 cuboids of 5 cm, 2 cm and 5 cm are required to form a cube.

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