Show that automorphism Aut(N/$\mathbb{Q}$) is isomorphic to symmetric group $S_3$ Let the extension $N/\mathbb{Q}$  be the splitting field of the of the polynomial $f (X) = X^3 - 2 ∈ Q [X]$, where $N ⊂ \mathbb{C}$. Show that:
The automorphism group Aut$(N / \mathbb{Q})$ is isomorphic to the symmetric group $S_3$.
Im currently studying for my algebra exam and this one exercise of an old algebra exam is rather hard. I really don't know where to begin, so maybe someone could give me hint? Thanks in advance!
Edit: Thanks for the quick answer! Ok, so the splitting field should be:
The polynom has the  zero points $\sqrt[3]{2}$, $\sqrt[3]{2}\zeta_3$ and $\sqrt[3]{2}\zeta_3^2$ with $\zeta_3=\frac{-1+i\sqrt{3}}{2}$. The splitting field of $f$ over $\mathbb{Q}$ should be $L=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$
But now I am not sure how to determine $\mathbb{Q}$-conjugates of $N$
 A: Your edit is very much on the right direction. The splitting field is the smallest field containing $\sqrt[3]{2}\zeta_3^j$ for $j = 0,1,2$. This is certainly contained in $\Bbb Q(\sqrt[3]{2}, \zeta_3)$ but since $\zeta_3 = (\sqrt[3]{2}\zeta_3^2) / (\sqrt[3]{2}\zeta_3^2)$ we have the other inclusion, hence the equality.
Now, this extension is Galois and it sits in a tower $N = \Bbb Q(\sqrt[3]{2})(\zeta_3) - \Bbb Q(\sqrt[3]{2}) - \Bbb Q$.
Hint: prove that
$$
[N:\Bbb Q] = [N : \Bbb Q(\sqrt[3]{2})][\Bbb Q(\sqrt[3]{2}) : \Bbb Q] = 6.
$$
Indeed,

 Since $X^2+X+1 \in \Bbb Q[X] \subset \Bbb Q(\sqrt[3]{2})[X]$ vanishes at $\zeta_3$ and $\zeta_3 \in \Bbb C\setminus \Bbb R$, we have $[N : \Bbb Q(\sqrt[3]{2})] = 2$. On the other hand since $X^3-2$ is irreducible over $\Bbb Q$ by Eisenstein's criterion, we have $m(\sqrt[3]{2},\Bbb Q) = X^3-2$ and $[\Bbb Q(\sqrt[3]{2}) : \Bbb Q] = 3$.

Hint': This gives $[N:\Bbb Q] = 6$ and so $G = \mathbf{Gal}(N/\Bbb Q)$ has order $6$.
If you already know that a splitting field of a polynomial of degree $n$ embeds as a subgroup of $S_n$, then you are done. Otherwise, since groups of order $6$ are characterized, analize them and describe some explicit automorphisms to conclude.

 There are only two such groups up to isomorphism, namely, $S_3$ and $\Bbb Z_6$. It suffices to show that $G$ is not $\Bbb Z_6$. One possible argument is to note that the maps defined by $\tau(\zeta_3) = \zeta_3^2, \tau(\sqrt[3]{2}) = \sqrt[3]{2}$ and $\sigma(\zeta_3) = \zeta_3, \sigma(\sqrt[3]{2}) = \sqrt[3]{2}\zeta_3$ are a transposition and $3$-cycle respectively who do not commute.

