Question: Prove that for any nonnegative integer $n$, $n^3 - n$ is divisible by $3$.  So I suppose that $n^3 - n = 3m$ for some integer $m$.  I know it is true for $n= 1$.  Suppose it is true for $n=k$. So $k^3 - k = 3t$ for some integer $t$.  How to prove it for $n= k+1$? I didn't get the equation for $n= k+1$.
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?
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