$a$ is a natural number that isn't equal to 0
My attempt:
$a(a^{2}-1)$ = $a(a-1)(a+1)$
There are 2 cases here:
If a = 2k:
We have $2k(2k+1)(2k+1)$ because $a-1$ and $a+1$ come before and after $a$ so if $a$ is even then they are both odd.
By calculation and simplification, we find this:
$8k^{3} + 8k^{2} + 2k$ which is useless because it still doesn't prove that it's divisible by 6
If a = 2k +1:
We have $(2k+1)(2k)(2k)$
By calculation and simplification, we find this: $8k^{3} + 4^{2}$ which is also nothing.
Basically what I wanted, in the end, is something of the form $6k$ or at least $3k$ and $2k$ both so that I can say it's divisible by 6, however, I can only prove that they are divisible by 2 which is not enough.
Any help with this?
