$x^n-dy^n$ irreducible?

Question

I am dealing right now with a generalized form of the Pell Equation. In order to use Thue's Theorem I need to know that $x^n-dy^n$ ist irreducible over $\mathbb{Q}$. Somehow I don't get why. I tried to use Eisenstein over $\mathbb{Q}[X]$ but that didn't work...

Answer 1

Suppose the polynomial were reducible. Then it would be reducible after substituting $y=1$. But now you can apply Eisenstein, since $d$ is assumed squarefree.

Answer 2

If it was reducible then $ (x/y)^n-d$ would be reducible too. Replace $x/y$ by $z$ and apply Eisenstein.