Prove that $n^3 - n$ is divisible by 6 by factoring I need to prove that $n^3 - n$ is divisible by $6$ by factoring it and by knowing that the product of each consecutive $3$ numbers is divisible by $2$ and $3$. I tried:
$n(n^2 - 1)$ 
Factoring it didn't really get me anywhere. I think I'm supposed to do something like:
$n[(n - 1)(n - 2)]$ And if I do the math in this I get back: 
$n[n^2 - 3n + 2] = n^3 - 3n^2 + 2n$ 
which looks like a dead end, unless I can somehow prove that:
$n^3 - n = n^3 - 3n^2 + 2n$ 
But that's impossible, since it's not true that $2n^2 = 2n$ for every positive integer
So what else can I do? These are the instructions the book gave - factor it and remember that the product of three consecutive number is divisible by 6.
 A: You were on the right track initially.
$$n^3-n = n(n^2-1) = n(n+1)(n-1).$$
These are three consecutive integers, so at least one of them is even. Additionally, $3$ must divide one of the numbers by the same logic. Thus $n^3-n$ is divisible by $6$ since it is divisible by $2$ and $3$.
A: Hint:$(n^2-1)=(n+1)(n-1)$, so
$$ n(n^2-1) = (n-1)n(n+1)$$
And because $6 = 2 \cdot 3 \wedge \gcd(2,3)=1$
$$6 \mid n \Longleftrightarrow 2 \mid n \wedge 3 \mid n $$
You should note too, that one of two numbers has to be even.
$$(\forall n \in \mathbb{Z})(2 \mid n~ \dot\vee ~2 \mid (n+1))$$
And similarly you can show divisible by three ($3$ has to divide one of three consecutive numbers).
If you want long explanation you can say: the remainder ($n \mod 3$) is $0$, then $3\mid n$, otherwise the remainder can be $2$ and $3 \mid (n+1)$ or $1$ and then $3 \mid (n+2)$. So 
$$(\forall n \in \mathbb{Z})\left(3 \mid (n-1)~ \dot\vee ~ 3 \mid n~ \dot\vee ~3 \mid (n+1)\right)$$

Pursuant to above claims.
$$\begin{split}
(2 \mid n ~\dot\vee~ 2 \mid (n+1))\wedge (3 \mid (n-1) ~\dot\vee~ (3 \mid n) ~\dot\vee~ 3 \mid (n+1)) &\Longleftrightarrow\\
\left(2 \mid n(n^2-1) \wedge 3 \mid n(n^2-1)\right)&\Longleftrightarrow 6 \mid n(n^2-1)
\end{split}$$
$\mathscr{Q.E.D.}$
A: $$N^3-N = N \cdot (N^2-1) = N \cdot (N-1) \cdot (N+1)$$
$N-1, \;N$ and $N+1$ have to be three consecutive numbers as $N$ is constant.
If you have three numbers, at least one of them has to have a factor of $2$ and one of them has a factor of $3$.
Since both $2$ and $3$ are factors, $\operatorname{lcm}(2,3)=6$ is also a factor.
