Question

$8640$ is not a perfect cube.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A True

Prime factorization of:

$8640=2^6\times 3^3\times 5^1$.

We know, a perfect cube has multiples of $3$ as powers of prime factors.

Here, $8640$ is not a perfect/complete cube.

That is, the given statement is true.

Hence, option $A$ is correct.