Question

Is $\mathrm{162,000}$ a perfect cube? If not, by which smallest natural number should $\mathrm{162,000}$ be divided so that the quotient is a perfect cube?

Answer 1

$\mathrm{162,000}=6\times 5\times 5\times 5\times 6\times 6\times 2\times 3$
The primes $2$ and $3$ do not appear in groups of three. So, $\mathrm{162,000}$ is not a perfect cube.
In the factorization of $\mathrm{162,000}$ the prime $2$ appears only one time and the prime $3$ appears once.
So, if we divide $\mathrm{162,000}$ by $2\times 2\times 3\times 3=36$, then the prime factorisation of the quotient will not contain $2$ and $3$.
Hence, the smallest natural number by which $\mathrm{162,000}$ should be divided to make it a perfect cube is $36$.