Is $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ a Galois extension of $\mathbb{Q}$

From a previous question, we have that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$, and I am assuming that we would need to use this to justify the answer to the question.

Would it be right to use some idea of a splitting field here? Or is it something to do with $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ not containing a primitive 3rd root of unity?

This is a past exam paper question worth 8 marks, so it doesn't seem like it's a few lined answer, but I am a little stuck on where to go.

Solution 1:

The extension $\;\Bbb Q(\sqrt[3]3,\,\sqrt[4]3)/\Bbb Q\;$ is real , yet the minimal polynomial of $\;\sqrt[3]3\;$ over $\;\Bbb Q\;$ , namely $\;x^3-3\;$, has complex non-real roots, which means the extension cannot be normal and thus neither Galois.