(i) The prime factorisation of 100 is as follows.
100 = 2 × 2 × 5 × 5
Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grouping the triplets. Therefore, 100 is not a perfect cube.
(ii) 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
Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.