Find the smallest number by which 135 must be divided to obtain a perfect cube.

$135=\underline{3\times 3\times 3}\times 5$

The prime factor 5 does not appear in a group of three. So, 135 is not a perfect cube. In the factorisation 5 appears only one time. If we divide the number by 5, then it will become a perfect cube.

So, $135÷5=\underline{3\times 3\times 3}=27$ is a perfect cube.
Hence, the smallest natural number by which 135 should be divided to make it a perfect cube is 5.