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 = 2
3 ☓ 3
3
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 = 2
3 ☓ 5
3
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.
∴ 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
= 2
3 ☓ 2
3 ☓ 3
3 ☓ 3
3
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.