If n is any natural number, then n ^ 3-N must have a divisor of?

If n is any natural number, then n ^ 3-N must have a divisor of?

n^3-n=(n-1)n(n+1)
One of three consecutive integers must be divisible by 3, and there must be at least one even number,
So n ^ 3-N must be a multiple of 6