Galois groups of $x^3-3x+1$ and $(x^3-2)(x^2+3)$ over $\mathbb{Q}$

Solution 1:

You do not need to know the roots of the cubic to find its Galois group. You should consult an algebra book about Galois groups and discriminants here. Then the solution is as follows. By the Rational Root Theorem you know that $x^3-3x+1$ is irreducible and its discriminant is $81$, which is a square in $\mathbb{Q}$. Therefore the Galois group of $x^3-3x+1$ is the alternating group $A_3$. If the discriminant of a cubic is not a square, and the polynomial is irreducible, then its Galois group is $S_3$. This is the case, for example, for $x^3+3x+1$.

Solution 2:

For the second one, you should think about those polynomials separately. In that case, it's pretty easy to see that the splitting field of $(x^3 - 2)(x^2+3)$ is $\mathbb{Q}(\sqrt[3]{2} , \zeta_3, \sqrt{-3})$.

Since $\zeta_3 = \frac{-1 \pm \sqrt{-3}}{2}$, this is the same as $\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3})$. Since $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt{-3}) : \mathbb{Q}] = 2$, and $(2,3) =1$, we get $[\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3}) :\mathbb{Q}] = 6$.

So, $Gal(\mathbb{Q}(\sqrt[3]{2} , \sqrt{-3})/\mathbb{Q})$ is either $\mathbb{Z}/6\mathbb{Z}$ or $S_3$. But, since the subfield $\mathbb{Q}(\sqrt[3]{2})$ is not Galois over $\mathbb{Q}$, the Galois group cannot be abelian, and so it is $S_3$.