## 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 the part of NCERT Solutions for Class 8 Maths (Rationalised Contents). Here you can
find the NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots
Ex 6.1**.**

### Ex 6.1 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) Let us find the prime factorization of 216:

216 = __2 × 2 × 2__ × __3 × 3 × 3__

Here, 2 and 3 form a group of three.

Thus, 216 is a perfect cube.

128 =

__2 × 2 × 2__×

__2 × 2 × 2__× 2

Here, 2 is left without making a group of three.

Thus, 128 is not a perfect cube.

(iii) Let us find the prime factorization of 1000:

1000 =

__2 × 2 × 2__×

__5 × 5 × 5__

Here, 2 and 5 form a group of three.

Thus, 1000 is a perfect cube.

(iv) Let us find the prime factorization of 100:

100 =

__2 × 2__×

__5 × 5__

Here, 2 and 5 do not form a group of three.

Thus, 100 is not a perfect cube.

(v) Let us find the prime factorization of 46656:

46656 =

__2 × 2 × 2__×

__2 × 2 × 2__×

__3 × 3 × 3__×

__3 × 3 × 3__

Here, 2 and 3 form the groups of three.

Thus, 46656 is a perfect cube.

### Ex 6.1 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) Let us find the prime factorization of 243:

243 = __3 × 3 × 3__ × 3 × 3 = 3^{3} × 3 × 3

Here, one more 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3.

Thus, the required smallest number to be multiplied is 3.

256 =

__2 × 2 × 2__×

__2 × 2 × 2__× 2 × 2 = 2

^{3}× 2

^{3}× 2 × 2

Here, one more 2 is required to make 2 × 2 a group of three, i.e., 2 × 2 × 2.

Thus, the required smallest number to be multiplied is 2.

72 =

__2 × 2 × 2__× 3 × 3 = 2

^{3}× 3 × 3

Here, one more 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3.

Thus, the required smallest number to be multiplied is 3.

(iv) Let us find the prime factorization of 675:

675 =

__3 × 3 × 3__× 5 × 5 = 3

^{3}× 5 × 5

Here, one more 5 is required to make 5 × 5 a group of three, i.e., 5 × 5 × 5.

Thus, the required smallest number to be multiplied is 5.

(v) Let us find the prime factorization of 100:

100 = 2 × 2 × 5 × 5

Here, the numbers 2 and 5 are required to make 2 × 2 × 5 × 5 a group of three, i.e., 2 × 2 × 2 × 5 × 5 × 5.

Thus, the required smallest number to be multiplied is 2 × 5 = 10.

### Ex 6.1 Class 8 Maths Question 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) 92

(v) 704

Solution:

(i) Let us find the prime factorization of 81:

81 = __3 × 3 × 3__ × 3 = 3^{3} × 3

Here, the number 3 is not in group of three. Thus, 3 is the number by which 81 is divided to make it a perfect cube, i.e., 81 ÷ 3 = 27, which is a perfect cube.

Thus, 3 is the required smallest number.

128 =

__2 × 2 × 2__×

__2 × 2 × 2__× 2 = 2

^{3}× 2

^{3}× 2

Here, the number 2 is not in group of three. Thus, 2 is the smallest number by which 128 is divided to make it a perfect cube, i.e., 128 ÷ 2 = 64, which is a perfect cube.

Thus, 2 is the required smallest number.

(iii) Let us find the prime factorization of 135:

135 =

__3 × 3 × 3__× 5 = 3

^{3}× 5

Here, the number 5 is not in group of three. Thus, 5 is the smallest number by which 135 is divided to make a perfect cube, i.e., 135 ÷ 5 = 27, which is a perfect cube.

Thus, 5 is the required smallest number.

(iv) Let us find the prime factorization of 192:

192 =

__2 × 2 × 2__×

__2 × 2 × 2__× 3 = 2

^{3}× 2

^{3}× 3

Here, the number 3 is not in group of three. Thus, 3 is the smallest number by which 192 is divided to make it a perfect cube, i.e., 192 ÷ 3 = 64, which is a perfect cube.

Thus, 3 is the required smallest number.

(v) Let us find the prime factorization of 704:

704 =

__2 × 2 × 2__×

__2 × 2 × 2__× 11 = 2

^{3}× 2

^{3}× 11

Here, the number 11 is not in group of three. Thus, 11 is the smallest number by which 704 is divided to make it a perfect cube, i.e., 704 ÷ 11 = 64, which is a perfect cube.

Thus, 11 is the required smallest number.

### Ex 6.1 Class 8 Maths Question 4.

### Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?

Solution:

The sides of the cuboid of plasticine are 5 cm, 2 cm and 5 cm.

Volume of the cuboid = 5 cm × 2 cm × 5 cm = 50 cm^{3}

Let us find the prime factorization of 50:

50 = 2 × 5 × 5

To make it a perfect cube, we must have

2 × 2 × 2 × 5 × 5 × 5

= 20 × (2 × 5 × 5)

= 20 × volume of the given cuboid

Thus, the required number of cuboids is 20.

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