For any integer, $n$, show that $n^3 \equiv 0$ or $\pm 1(\mod 7)$. Use theory of congruences
So I thought about a couple of ways to go with this. I thought about showing $7|n^3$ or $7|n^3\pm1$ to be true by letting $n=7k, 7k\pm1, 7k\pm2, 7k\pm3$ and proving it that way but that seemed like a long route. Also, the problem asks to use theory of congruences and I'm not sure if this approach really uses that.
Is there a way to use the fact that $a^n \equiv b^n (\mod m)$? I tried following this logic, but I got stuck trying to imply that $n \equiv 0$ or $\pm 1(\mod 7)$ by removing the cube root.
Any help is appreciated!