I agree it should be there in the question.

So, in other words, is there a natural number n such that sum of cubes of first n natural numbers equals m?

essentially what the question is asking you is:

following the sequence n^3 + (n-1)^3 + (n-2)^3 ... 1^3, is there a value, n, you can put in such that the summation of the sequence == m (m is the parameter).

if there is none, return -1. Your work is to find n (if there is).

.....
your first two tests are correct. three is also correct, but you don't have to repeat.