Galois Group Solve for radicals Let $f(x)=x^3+px+q\in \mathbb{Q}\left[x\right]$ and $K\subseteq \mathbb{C}$ a descomposition field of $f$. Let $\alpha_1,\alpha_2,\alpha_3$ the roots of $f$ in $K$. Suppose $\alpha_1\in \mathbb{R}$ and $\alpha_3=\overline{\alpha_2}$ where   $\alpha_3,\overline{\alpha_2}\in \mathbb{C}.$ Show that $\text{Gal}(K/\mathbb{Q})\cong S_3.$
Hello! I don't understand this problem.. How can solve this?

I assume $f$ is irreducible over $\mathbb{Q}$ because otherwise the statement is false. Take $x^3+x=x(x-i)(x+i)$ as a counterexample. 
I will use a known lemma from group theory. 
Lemma: If $p$ is a prime and $H\leq S_p$ is a subgroup which contains a transposition and an element of order $p$ then $H=S_p$. 
Now back to your question. Easy to see that the restriction of the complex conjugation to $K$ is an element in $Gal(K/\mathbb{Q})$ and the corresponding permutation is $(\alpha_2\ \alpha_3)$. So $Gal(K/\mathbb{Q})$ contains a transposition. Also, $\mathbb{Q}(\alpha_1)$ is an intermediate field and hence $[\mathbb{Q}(\alpha_1):\mathbb{Q}]$ divides $[K:\mathbb{Q}]=|Gal(K/\mathbb{Q})|$. Since $f$ is irreducible we know it is the minimal polynomial of $\alpha_1$ and hence $[\mathbb{Q}(\alpha_1):\mathbb{Q}]=3$. So $|Gal(K/\mathbb{Q})|$ is divisible by $3$ and by Cauchy's theorem it contains an element of order $3$. By the lemma we conclude that $Gal(K/\mathbb{Q})\cong S_3$. 
