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:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$

You can decompose $n^3-n = (n-1)n(n+1)$. So $n^3-n$ is a product of three consecutive integers, and in such a sequence you always have an even number and a multiple of $3$. Hence $n^3-n$ is a multiple of $6$.
By induction: the base case is $n = 0$, which is trivial. Now suppose $6$ divides $n^3-n$. Claim: $6$ divides $(n+1)^3-(n+1)$.
You have
$$(n+1)^3-(n+1) =n^3+3n^2+3n+1-n-1=n^3-n+3n(n+1).$$
Now observe that since $6$ divides $n^3-n$ you just need to show that $3n(n+1)$ is a multiple of $6$. But one between $n$ and $n+1$ is even, so $3n(n+1)$ is a multiple of $6$. Hence $6$ divides $(n+1)^3-(n+1)$.
