(i) Prime factorization of 216:

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³

Here we can see that each prime factor is appearing as many times as a perfect multiple of 3, hence, 216 is a perfect cube.

(ii)The prime factorization of 128 is:

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, we can observe that each prime factor is not appearing as many times as a perfect multiple of 3.

One 2 is remaining after grouping the triplets of 2 Hence, 128 is not a perfect cube.

(iii) The prime factorization of 1000 is:

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, we can observe that each prime factor is appearing as many times as a perfect multiple of 3.

Hence, 1000 is a perfect cube

(iv)The prime factorization of 100 is:

100 = 2 × 2 × 5 × 5

Here, we can see that every prime factor is not appearing as many times as a perfect multiple of 3.

Two 2s and two 5s are remaining if we group the triplets.

Therefore, 100 is not a perfect cube

(v)The prime factorization of 46656 is:

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, we can see that number of 2s and 3s is 6 each and we know that 6 is divisible by 3.

Hence, 46656 is a perfect cube