Galois Group of $\mathbb{Q}(\sqrt[3]{3},i\sqrt{3},\zeta_{13})/ \mathbb{Q}$ I would like find the Galois Group of 
$$\mathbb{Q}(\sqrt[3]{3},i\sqrt{3},\zeta_{13})/ \mathbb{Q},$$
that field extension is Galois because is the splitting field of separable polynomial $(x^3-3)(x^{13}-1)$. We know $[\mathbb{Q}(\sqrt[3]{3},i\sqrt{3}):\mathbb{Q}]=6$ and $[\mathbb{Q}(\zeta_{13}):\mathbb{Q}]=12$, so 
$$[\mathbb{Q}(\sqrt[3]{3},i\sqrt{3},\zeta_{13}): \mathbb{Q}]\leq 72.$$
Note that $\text{Gal}(\mathbb{Q}(\sqrt[3]{3},i\sqrt{3})/\mathbb{Q}) \cong S_3$ is non abelian, so that, implies that 
$\sqrt[3]{3} \notin \mathbb{Q}(\zeta_{13})$, because $\mathbb{Q}(\zeta_{13})$ is an abelian Galois field extension (isomorphic to $\mathbb{Z}_{13}^{*})$. We can show that $i\sqrt{3} \notin \mathbb{Q}(\zeta_{13})$ because the only quadratic extensión of $\mathbb{Q}(\zeta_{13})$ is a real extension. ¿which is the degree of the galois extension?

Note that $\mathbb{Q}(\sqrt{-3},\zeta_{13})=\mathbb{Q}(\zeta_3,\zeta_{13})=\mathbb{Q}(\zeta_{39})$ and the only degree $3$ extension of $\mathbb{Q}$ comes from $\mathbb{Q}(\zeta_{13})$, so the same argument applies to give you $[\mathbb{Q}(\zeta_{39})(\sqrt[3]3):\mathbb{Q}]=72$.
The Galois group is $S_3\times C_{12}$.
