CBSE Class 8 - Maths - Cube and Cube Roots (Ex 7.1)

Cube and Cube Roots 

(NCERT Ex 7.1)

Q1: 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 = 2³ ☓ 3³

Here, each prime factor appears as many times as a perfect multiple of 3, therefore, 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

Here, each prime factor does not appear as many times as a perfect multiple of 3. Therefore, 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 =  2³ ☓ 5³

Here, as each prime factor appears as many times as a perfect multiple of 3, therefore, 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

Here, each prime factor does not form triplet group(s). Therefore, 100 is not a perfect cube.

(iv) 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
 = 2³ ☓ 2³ ☓ 3³ ☓ 3³
Here, each prime factor appears as many times as a perfect multiple of 3,
∴ 46656 is a perfect cube.

Q2: 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) 243 = 3 ☓ 3 ☓ 3☓ 3 ☓ 3

Here, two 3s are left which require one more 3 to form a triplet.
i.e. 243 ☓ 3 = 3 ☓ 3 ☓ 3 ☓ 3 ☓ 3 ☓ 3 = 729 is a perfect cube.

∴ the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) 256 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2

Here, two 2s are left which requires one more 2 to form a triplet. To make 256 a cube, one more 2 is required.

256 ☓ 2 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2☓2 ☓ 2 ☓ 2 = 512 is a perfect cube.

Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

(iii) 72 = 2 ☓ 2 ☓ 2 ☓ 3 ☓ 3

Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required.

72 ☓ 3 = 2 ☓ 2 ☓ 2 ☓ 3 ☓ 3 ☓ 3 = 216 is a perfect cube.

Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3.

(iv) 675 = 3 ☓ 3 ☓ 3 ☓ 5 ☓ 5

Here, two 5s are left which require one more 5 to forma a triplet.

∴ 675 ☓ 5 = 3 ☓ 3 ☓ 3 ☓5 ☓ 5 ☓ 5 = 3375 is a perfect cube.

Hence, the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5.

(v) 100 = 2 ☓ 2 ☓ 5 ☓ 5

Here, two 2s and two 5s are left which are do not form a triplet. To make 100 a cube, we need one more 2 and one more 5.

∴ 100 ☓ 2 ☓ 5 = 2 ☓ 2 ☓ 2 ☓ 5 ☓ 5 ☓ 5 = 1000 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.

Q3: 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) 81 = 3 ☓ 3 ☓ 3 ☓ 3

Here, one 3 is left which does not form a triplet.

If we divide 81 by 3, then it will become a perfect cube.

∴ 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) 128 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2

Here, one 2 is left which is not in a triplet.

If we divide 128 by 2, then it will become a perfect cube.

∴ 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) 135 = 3 ☓ 3 ☓ 3 ☓ 5

Here, one 5 is left which is not part of a triplet.

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) 192 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 3

Here, one 3 is left which is not part of a triplet.

By dividing 192 by 3, it becomes 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) 704 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ 11

Here, one 11 is left which does not form a triplet.

If we divide 704 by 11, then it makes 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.


Q4: 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 cube 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 do not form a triplet.

To make it a pefect cube, multiply this expression by 2 ☓ 2 ☓ 5 = 20, then it will become a perfect cube.

i.e. (5 ☓ 5 ☓ 2 ☓ 2 ☓ 2 ☓ 5) = (5 ☓ 5 ☓ 5 ☓ 2 ☓ 2 ☓ 2) = 1000 is a perfect cube.

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