Determine the splitting field and its degree over $\mathbb{Q}$ for $x^6 - 4$.

\subsubsection*{Solution}

We have
where naturally $k \in { 0, 1, 2 }$. Thus $x^6 - 4$ splits completely in $\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$. Since $x^3 - 2$ and $x^2 + 3$ are both irreducible in $\mathbb{Q}$ by Eisensteinâ€™s criterion, the degree of this extension over the rationals is $2 \times 3 = 6$.