Show that $n^3-n$ is divisible by $6$ using induction As homework, I have to prove that

$\forall n \in \mathbb{N}: n^3-n$ is divisible by 6

I used induction
1) basis: $A(0): 0^3-0 = 6x$ , $x \in \mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?
2) requirement: $A(n):n^3-n=6x$
3) statement: $A(n+1): (n+1)^3-(n+1)=6x$
4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$
So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $\forall n \in \mathbb{N}$
Did I do something wrong or is it that simple?

No need for induction.
$$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
which are three consecutive integers. So one must be divisible by 3.

Check for $n=1$: $1^3-1=0=3\cdot 0$
Assume it's true for $n=k$.
If you let $n=k+1$ you get
$$\begin{align*}
(k+1)^3-(k+1)&=k^3+3k^2+2 \\
&=k^3+3k^2+2k\\
&=3\cdot (k^2+k)+(k^3-k)\end{align*}$$
 which is divisible by 3
