Prove that $\sqrt[3]p+\sqrt[3]{p^5}$ is irrational if $p$ is prime

Question

Prove that $\sqrt[3]p+\sqrt[3]{p^5}$ is irrational when $p$ is a prime.

First I suppose $x=\sqrt[3]p+\sqrt[3]{p^5}$. Cubing gives $$x^3=p+p^5+p^2x$$ And then what properties of prime, and how to test its irrationallity?

Answer 1

By Eisenstein's criterion, $X^3-p$ is irreducible over $\Bbb Q$. Therefore $1$, $\sqrt[3]p$ and $\sqrt[3]{p^2}$ are linearly independent over $\Bbb Q$.

Now observe that $x=\sqrt[3]p+p\sqrt[3]{p^2}$.