Find all values that make the expression a perfect cube Find all the positive integers $n$ such that $n^3-n$ is a perfect cube.

If $n > 1$, then $3n^2-4n+1 = (3n-1)(n-1) > 0$. 
Hence, $(n-1)^3 = n^3-3n^2+3n-1 < n^3-n < n^3$. 
Since $n^3-n$ is strictly between two consecutive perfect cubes, $n^3-n$ is not a perfect cube. 
The only remaining positive integer is $n = 1$, which yields $n^3-n = 0 = 0^3$. 
Therefore, the only positive integer $n$ such that $n^3-n$ is a perfect cube is $n = 1$.
