elementary number theory - Proof that $sqrt[3]{17}$ is irrational

Consider $\sqrt[3]{17}$. Like the famous proof that $\sqrt2$ is irrational, I also wish to prove that this number is irrational. Suppose it is rational, then we can write:

$$17 = \frac{p^3}{q^3}.$$
and then

$$17q^3 = p^3$$

With the proof of $\sqrt2$ we used the fact that we got an even number at this step in the proof and that $p$ and $q$ were in lowest terms. However, 17 is a prime number, somehow we could use this fact and the fact that every number has a unique prime factorisation to arrive at a contradiction, but I don't quite see it yet.

Answer

The argument that works with $2$ also works with $17$. Since $17q^3=p^3$, $17\mid p^3$ and therefore $17\mid p$. Can you take it from here?