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**ThePerfectHacker** This irreducible polynomial has 1 real root and 2 complex roots that are complex conjugates. If $\displaystyle \tau$ is complex conjugation then we see that $\displaystyle \tau$ is a transposition (2-cycle) if viewed as an element of $\displaystyle S_3$ (remember that the Galois group is a subgroup of $\displaystyle S_3$). Now if $\displaystyle \alpha$ is a root of $\displaystyle f$ then $\displaystyle \mathbb{Q}(\alpha)/\mathbb{Q}$ is a degree $\displaystyle 3$ extension and so $\displaystyle 3$ divides the degree of the splitting field over $\displaystyle \mathbb{Q}$. Thus, $\displaystyle 3$ divides the order of the Galois group, hence there is an element of order 3 in the Galois group i.e. a 3-cycle. But any subgroup of $\displaystyle S_3$ generated by a 2-cycle and a 3-cycle is $\displaystyle S_3$ itself, so the Galois group is $\displaystyle S_3$.