I'm working on the problem:
Let $\zeta=e^{2\pi i/3}$, let $\alpha=\sqrt[3]{a+b\sqrt{2}}$ and let $L$ be the splitting field for an irreducible polynomial for $\alpha$ over $\mathbb{Q}(\zeta)$. Determine the possible Galois groups of $L$ over $\mathbb{Q}(\zeta)$.
For this I've eliminated the two easy cases. If $b=0$, then its either $C_1$ or $C_3$ depending whether $a$ is a third power. If $b\neq 0$, we have the quadratic subfield $\mathbb{Q}(\zeta, \sqrt{2})$. If $\alpha \in \mathbb{Q}(\zeta, \sqrt{2})$ then $L=\mathbb{Q}(\zeta, \sqrt{2})$ and $\text{Gal}(L/\mathbb{Q}(\zeta))=C_2$.
Finally, were left with the interesting case $b\neq 0$ and $\alpha\not\in \mathbb{Q}(\zeta, \sqrt{2})$. In this case we have the field diagram
$$ \hspace{-2em}L=\mathbb Q\left(\zeta_3,\sqrt[3]{a+\sqrt2b},\sqrt[3]{a-\sqrt2b}\right) \\ \phantom{\quad \text{1 or 3}}\Rule{1px}{2em}{2em} \quad \text{ 1 or 3} \\ \mathbb Q\left(\zeta_3,\sqrt[3]{a+\sqrt2b}\right) \\ \phantom{\quad \text{3}}\Rule{1px}{2em}{2em} \quad \text{ 3} \\ \mathbb Q(\zeta_3,\sqrt2) \\ \phantom{ \quad \text{2}}\Rule{1px}{2em}{2em} \quad \text{ 2} \\ \mathbb Q(\zeta_3) $$
Note, I know the splitting field because $\alpha \not\in \mathbb{Q}(\zeta, \sqrt{2})$ means $[\mathbb{Q}(\zeta, \alpha):\mathbb{Q}(\zeta,\sqrt{2})]=3$ since its a Kummer extension. Thus $[\mathbb{Q}(\zeta,\alpha):\mathbb{Q}(\zeta)]=6$ and
$$(X^3-a)^2-2b^2$$
which has roots $\pm\sqrt[3]{a\pm b\sqrt{2}}$ must be the minimal polynomial of $\alpha$ over $\mathbb{Q}(\zeta)$. Since $|\text{Gal}(L/\mathbb{Q}(\zeta))|=6\text{ or }18$ and its a transitive subgroup of $S_6$, googling shows it must be $C_6$, $S_3$, or $C_3\times S_3$, although the problem should be solvable by hand. The $C_6$ case is easy taking $\alpha =2^{1/6}$, but I get stuck on the $S_3$ case and degree $18$ case.
I would prefer just a hint so that I could try finishing it myself.
Edit: I think I have constructed the $S_3$ case. Take $\alpha=\sqrt[3]{1+\sqrt{2}}$. It's easy to check we fall into the interesting case and $\sqrt[3]{1-\sqrt{2}}=-\alpha^{-1}$ so $L$ is degree $6$. Let $\beta = \sqrt[3]{1-\sqrt{2}}=-\alpha^{-1}$. Any element of $\text{Gal}(L/\mathbb{Q}(\zeta))$ must send $\sqrt{2}\mapsto \pm\sqrt{2}$. Now $\alpha$ and $\beta$ satisfy
$$\alpha^3-(1+\sqrt{2})=0\qquad \beta^3-(1-\sqrt{2})=0.$$
Thus, suppose $\psi\in\text{Gal}(L/\mathbb{Q}(\zeta))$. If $\psi(\sqrt{2})=\sqrt{2}$, then $\psi(\alpha)\in\{\alpha,\zeta\alpha,\zeta^2\alpha\}$ and if $\psi(\sqrt{2})=-\sqrt{2}$, then $\psi(\alpha)\in\{\beta,\zeta\beta,\zeta^2\beta\}$. Since this gives at most $6$ automorphisms, each one must indeed be one. Thus consider the automorphisms
$$\sigma=\begin{cases}
\alpha \mapsto -\alpha^{-1} \\ \\
\sqrt{2}\mapsto -\sqrt{2}
\end{cases},\qquad
\tau = \begin{cases}
\alpha \mapsto \zeta\alpha \\ \\
\sqrt{2}\mapsto \sqrt{2}
\end{cases}.$$
We then have that
$$
\sigma\tau = \begin{cases}
\alpha \mapsto -\zeta\alpha^{-1} \\ \\
\sqrt{2}\mapsto -\sqrt{2}
\end{cases}\neq
\begin{cases}
\alpha \mapsto -\zeta^{-1}\alpha^{-1} \\ \\
\sqrt{2}\mapsto -\sqrt{2}
\end{cases} = \tau\sigma
$$
so $\text{Gal}(L/\mathbb{Q}(\zeta))$ is non-abelian and thus $\cong S_3$. I have confirmed this with Magma, but my proof could still be erroneous.
Edit 2: Here is a degree $18$ case. Take $\alpha=\sqrt[3]{1+1/\sqrt{2}}$. Then $L$ contains $\alpha$ and $\beta=\sqrt[3]{1-1/\sqrt{2}}$. Now
$$\alpha\beta=2^{-1/3}.$$
Thus $2^{1/2},2^{1/3}\in L$ so $2^{1/6}\in L$. We also then have
$$2^{1/6}\alpha = \sqrt[3]{1+\sqrt{2}},\;(\zeta + 1)2^{1/6}\beta=\sqrt[3]{1-\sqrt{2}}\in L.$$
Thus $\mathbb{Q}(\zeta, 2^{1/6}), \mathbb{Q}(\zeta, \sqrt[3]{1+\sqrt{2}}, \sqrt[3]{1-\sqrt{2}})\le L$, which by the above means $L$ must be degree $18$ since these two degree $6$ subextensions have different Galois groups over $\mathbb{Q}(\zeta)$. Finally, cheating a little bit, the only order $18$ transitive subgroup of $S_6$ is $C_6\times S_3$ so we know $\text{Gal}(L/\mathbb{Q}(\zeta))\cong C_6\times S_3$ and we have constructed the remaining case.
At this point I'm interested in people checking my proofs and/or giving any shorter answers, especially any that don't "cheat" by looking up the transitive subgroups of $S_6$, preferably without equivalently annoying group bashing, if possible.