Irreducibility of $x^5-9$

Question

Let $f:=X^5-9 \in \mathbb{Q}[X]$. Show: $f$ is irreducible. So I can't use the criterion, that f has no roots in $\mathbb{Q}[X]$, beacuse $\deg(f)>3$. Eisenstein is also not possible, because the only divider of $9$ is $3$. So I tried the reduction criterion, which doesn't really help me because $\deg(f)$ is still $5$. And in $\mathbb{Z}/7$ for example there are no prim elements, So I can't use Eisenstein. Is there another way I don't see?

Answer 1

The roots are $\sqrt[5]9\left(\cos(\frac {2k\pi}5+i\sin(\frac {2k \pi}5)\right)$ for $k \in [0,4]$ The trig functions have known values that involve $\sqrt 5$. Any product of two of them will still have the $\sqrt[5]{81}$ factor out front, so will not have rational coefficients.