With what least number should 8640 be divided so that the quotient is a perfect cube?

Finding prime factors of 8640

$8640 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)\times 5$

Since the prime factor 5 is not in triplets,

8640 will be divided by 5 so that the product is a perfect cube.