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 $n=0 \mod 3$ then is nothing to prove. 
Suppose that $(n,3)=1.$
Now it follows from the  Ferma litle theorem. We  have  that $n^3=n \mod 3$  or, in other words, $n^3-n$ is multiple of 3. 
