Prove that $f=X^3-\sqrt{3}$ is irreducible in $\mathbb{Q}(3^{1/4})$

Question

I tried to prove it by following that if $\alpha$ is a root of $f$, we can't express it using a basis of $\mathbb{Q}(3^{1/4})$. But I didn't get too far.

Answer 1

Let $\alpha$ be a root, meaning that $\alpha^6=3,$ so $\alpha$ is a root of $x^6-3$. Since the latter is irreducible in $\mathbb{Q}[x]$, $\alpha$ has degree $6$ over $\mathbb{Q}$, whereas the extension $\mathbb{Q}(3^{1/4})/\mathbb{Q}$ is of degree $4$, so $\alpha\not\in\mathbb{Q}(3^{1/4})$.