By what smallest number should 3600 be multiplied, so that the quotient is a perfect cube. Also, find the cube root of the quotient.
Prime factors of 3600 = 2x2x2x2x3x3x5x5
Grouping the factors into triplets of equal factors, we get
$3600=\underline{2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5}$
We know that, if a number is to be a perfect cube, then each of its prime factors must occur thrice.
We find that 2 occurs once 3 and 5 occurs twice only.
Hence, the smallest number, by which the given number must be multiplied in order that the product is a perfect cube $=2 \times 2 \times 3 \times 5=60$
Also, product $=3600 \times 60=216000$
Now, arranging into triplets of equal prime factors, we have
$216000=\underline{2 \times 2 \times 2} \times 2 \times 2 \times 2 \times \underline{3} \times 3 \times 3 \times 5 \times 5 \times 5$
Taking one factor from each triplets, we get
$\sqrt[3]{216000}=2 \times 2 \times 3 \times 5=60$