Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$

To prove by induction take the base case $n=0$ or $n=1$.
Then for the inductive step $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3(n^2+n)$$and the statement will be true if you can show that $n^2+n=n(n+1)$ is even. You can do that various ways, including a similar induction
