there are different cases to consider:
firstly, n^3-n can be expressed as: n(n-1)(n+1)
so, first, if n is divisible by 3, then n-1 is divisible by2, so whole thing divisible by 6 - nothing further needed there.
second thing you do is examine what happens when n is both odd and even: odd n=2k+1 for some k in Z - you find out, by factoring, that both expressions are divisible by 2,
3rd, you examine n when it`s NOT divisible by 3 - by using in the expression, (then factoring it) both n=3k+1, and n=3k-1 (this ensures thatyou encompass every number NOT divisible by 3) - you find that the expression n^3-1 when used with these values, simplifies to something with a factor of 3,
so you will have proved simultaneously that the expression has factors of both 2 and 3...