Galois group of $x^3+2x+1$

The polynomial \[ f(x)=x^3+2x+1 \] has no rational root, hence is irreducible. Its discriminant is \[ \Delta_f=-4(2)^3-27=-59, \] which is not a square in \(\mathbb Q\). Therefore the Galois group of its splitting field is \[ S_3. \] Since \(f\) is the minimal polynomial of \(\alpha\), \[ [\mathbb Q(\alpha):\mathbb Q]=3. \]

Now \[ g(x)=x^3+3x+1 \] also has no rational root, so it is irreducible of degree \(3\), and \[ \Delta_g=-4(3)^3-27=-135=-9\cdot15. \] Suppose \(g\) had a root \(\beta\in\mathbb Q(\alpha)\). Then \[ [\mathbb Q(\beta):\mathbb Q]=3=[\mathbb Q(\alpha):\mathbb Q], \] so \(\mathbb Q(\beta)=\mathbb Q(\alpha)\). Their normal closures over \(\mathbb Q\) would therefore coincide. For an irreducible cubic with Galois group \(S_3\), the unique quadratic subfield of its normal closure is generated by the square root of the discriminant. Hence we would have \[ \mathbb Q(\sqrt{-59})=\mathbb Q(\sqrt{-15}), \] which is false because \(59/15\) is not a rational square. Thus \(g\) has no root in \(\mathbb Q(\alpha)\); since it is cubic, it remains irreducible over \(\mathbb Q(\alpha)\).