**Question:**

Which of the following numbers is a perfect cube?

(a) 243

(b) 216

(c) 392

(d) 8640

**Solution:**

(b) For option (a) We have, 243

Resolving 243 into prime factors, we have

243= 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we get

243 = (3 x 3 x 3) x 3 x 3

Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 3 x 3.

Therefore, 243 is not a perfect cube.

For option (b) We have, 216 Resolving 216 into prime factqrs, we have

216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we get 216 = (2 x 2 x 2) x (3 x 3 x 3)

Clearly, in grouping, the factors of triplets of equal factors, no factor is left over.

So, 216 is a perfect cube.

For option (c) We have, 392

Resolving 392 into prime factors, we get

392 = 2 x 2 x 2 x 7 x 7

Grouping the factors in triplets of equal factors, we get

392 = (2 x 2 x 2) x 7 x 7

Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 7 x 7.

Therefore, 392 is not a perfect cube.

For option (d) We have, 8640

Resolving 8640 into prime factors, we get

8640=2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5

Grouping the factors in triplets of equal factors, we get

8640 = (2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3) x 5

Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor 5. Therefore, 8640 in not a perfect cube.

After solving, it is clear that option (b) is correct.