Finding Galois group of $x^6 - 3x^3 + 2$ I'm trying to find the Galois group of $$f(x)= x^6 - 3x^3 + 2$$ over $\mathbb{Q}$. 
Now I can factorise this as $$f(x) = (x-1)(x^2 + x + 1)(x^3 - 2)$$
I can see the splitting field must be $\mathbb{Q}(\omega, \sqrt[3]{2})$ (where $\omega$ is a 3rd root of unity) which has degree $6$ and so I know the Galois group must be order $6$. Now how do I distinguish between the cases $S_3$ or $C_6$?
I'm pretty sure it's $S_3$ but I'm having trouble seeing what the automorphisms actually are. I know that an element $\sigma$ in the Galois group must permute the roots of each irreducible factor and I also know that an automorphism is determined by it's action on $\sqrt[3]{2}$ and $\omega$ but I'm having trouble seeing what the actual automorphisms are in this case? 
E.g. I can see sending $\omega \to \omega^2$ permutes the roots of $x^2 + x + 1$, but then what do I do to $\sqrt[3]{2}$? Can that be left fixed? 
Thanks for any help!
 A: Let $\alpha := \sqrt[3]{2}$. As you said, any automorphism $\sigma \in Aut_{\mathbb{Q}} L $ is complete determined by 
$$\sigma (1) \in \{1\}\\\sigma(\omega) \in \{\omega,\omega^2\}\\\sigma(\alpha) \in \{\alpha, \alpha \omega,\alpha \omega^2\}*$$ 
Now set the table of possibilities, to find all automorphisms.
$(*)$ Notice that $\alpha, \alpha \omega,\alpha \omega^2$ are the roots of $X^3-2$.  
A: A basis for the splitting field of $x^6 - 3x^3 + 2$ over $\mathbb{Q}$ is $\{1, \omega, \sqrt[3]{2}, \omega\sqrt[3]{2}, \sqrt[3]{2}^2, \omega\sqrt[3]{2}^2\}$.
An element $\sigma \in \operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ is completely determined by $\sigma(\omega)$ and $\sigma(\sqrt[3]{2})$; note however that not every choice results in an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ (e.g. $\sigma(\omega) = \sigma(\sqrt[3]{2}) = 1)$.
As $\sigma \in \operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ permutes the  roots of minimal polynomials, we see that $\sigma(\omega) \in \{\omega, \omega^2\}$ and $\sigma(\sqrt[3]{2}) \in \{\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}\}$.
Let $\alpha(\omega) = \omega^2$ and $\alpha(\sqrt[3]{2}) = \sqrt[3]{2}$; $\alpha$ extends to an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. In particular
\begin{align*}
\alpha : 1 &\mapsto 1\\
\omega &\mapsto \omega^2\\
\omega^2 & \mapsto \omega\\
\sqrt[3]{2} &\mapsto \sqrt[3]{2}\\
\omega\sqrt[3]{2} &\mapsto \omega^2\sqrt[3]{2}\\
\omega^2\sqrt[3]{2} &\mapsto \omega\sqrt[3]{2}.
\end{align*}
Likewise, let $\beta(\omega) = \omega$ and $\beta(\sqrt[3]{2}) = \omega\sqrt[3]{2}$; $\beta$ extends to an element of $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. In particular
\begin{align*}
\beta: 1 &\mapsto 1\\
\omega &\mapsto \omega\\
\omega^2 & \mapsto \omega^2\\
\sqrt[3]{2} &\mapsto \omega\sqrt[3]{2}\\
\omega\sqrt[3]{2} &\mapsto \omega^2\sqrt[3]{2}\\
\omega^2\sqrt[3]{2} &\mapsto \sqrt[3]{2}.
\end{align*}
As $\alpha$ has order two and $\beta$ has order three, they generate $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$. By comparing $\alpha\circ\beta$ and $\beta\circ\alpha$, we can determine whether $\operatorname{Gal}(\mathbb{Q}(\omega, \sqrt[3]{2}), \mathbb{Q})$ is abelian or not, and hence whether it is $C_6$ or $S_3$.
