Proof that $n^3-n$ is a multiple of $3$. I'm struggling with this problem of proof by induction:
For any natural number $n$, prove that $n^3-n$ is a multiple of $3$.

I assumed that $k^3-k=3r$
I want to show that $(k+1)^3-(K+1)=3r$
The final statement is $K^3 +3K^2+2K$
Am I missing something ?

Use Fermat's little theorem.
Case1: if n is a multiple of 3 then trivially $n^3 -n$ is a multiple of $3$.
Case 2: If $n$ is not a multiple of $3$ you shall get $n^2 \equiv 1 \pmod 3$. Multiply $n$ and get $n^3 -n \equiv 0 \pmod 3$.
Now get your result.
