I need to find the minimal polynomial of $\sqrt[5]{3} \sqrt{2} i$ over $\mathbb{Q}$ and prove that this is the minimal polynomial. I let $x =\sqrt[5]{3} \sqrt{2} i$. Then $x^5 = 3 \sqrt{32}i$ and so $x^{10} + 288 = 0$. I want to prove this is the minimal polynomial now, by showing it is irreducible over $\mathbb{Q}$.
I consider the well known ring morphism $\phi : \mathbb{Q}[X] \to \mathbb{Z}_5[X]$ and consider the polynomial $x^{10} + 288 = x^{10} + 3 \in \mathbb{Z}_5[X]$. How can I conclude that this polynomial is irreducible in $\mathbb{Z}_5[X]$?