Compute the Galois group of the cubic polynomial \(f(x) = x^3 + 6x + 3\) over \(\mathbb{Q}\).
Solution
The polynomial \[ f(x)=x^3+6x+3 \] is Eisenstein at \(3\), hence irreducible over \(\mathbb Q\). Therefore its Galois group is a transitive subgroup of \(S_3\), so it is either \(A_3\) or \(S_3\).
For a depressed cubic \(x^3+px+q\), the discriminant is \[ \Delta=-4p^3-27q^2. \] Here \[ \Delta=-4\cdot6^3-27\cdot3^2=-1107=-3^3\cdot41, \] which is not a square in \(\mathbb Q\).
For an irreducible cubic, the Galois group lies in \(A_3\) exactly when the discriminant is a square. Hence \[ \boxed{\operatorname{Gal}(f/\mathbb Q)\cong S_3}. \] Equivalently, \(f'(x)=3x^2+6>0\), so \(f\) has one real root and one nonreal conjugate pair; complex conjugation is therefore a transposition in the transitive Galois group.