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We know that $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, and $[\mathbb{Q}(\sqrt[3]{2},i\sqrt{3}):\mathbb{Q}]=6$.

Now, consider an element $\alpha$ in the Galois group $\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2},i\sqrt{3}):\mathbb{Q})$. We know that $\alpha$ fixes $\mathbb{Q}$, while permuting the three roots of $x^3-2$. There are six possible permutations.

Question: How can we know that all six in fact belong to the Galois group? (Maybe some of them is not an automorphism of $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$.)

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$\newcommand{\size}[1]{\lvert #1 \rvert}$By the definition of the splitting field you have $$ E = \mathbb{Q}(\sqrt[3]{2},\sqrt{-3}) = \mathbb{Q}(r_{1}, r_{2}, r_{3}), $$ where $r_{1}, r_{2}, r_{3}$ are the roots of $f = x^3-2$.

Thus an element $\alpha \in G = \mathrm{Gal}(E:\mathbb{Q})$ is completely determined by the permutation it induces on the set $\Omega = \{ r_{1}, r_{2}, r_{3} \}$.

(More formally, the map $$ \Xi : G \to S_{\Omega}, \qquad \alpha \mapsto \alpha|_{\Omega} $$ is an injective homormorphism.)

Now $E/\mathbb{Q}$ is a Galois extension, so $\size{G} = \size{E:\mathbb{Q}} = 6$.

Since $G$ has six elements, it must induce six distinct permutations of $\Omega$, that is, all possible ones.

(More formally, $\Xi$ is an injective map between finite sets of the same size, so it is also onto.)

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