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 ?

If you want induction, here you go :
$k^3-k=3m$
$(k+1)^3-(k+1)=k^3+3k^2+3k+1-k-1=(k^3-k)+3(k^2+k)=3m+3n=3(m+n)$
This should help you i guess..
