TITLE: Galois groups of $x^3-3x+1$ and $(x^3-2)(x^2+3)$ over $\mathbb{Q}$
QUESTION [8 upvotes]: I want to find the Galois groups of the following polynomials over $\mathbb{Q}$. The specific problems I am having is finding the roots of the first polynomial and dealing with a degree $6$ polynomial.

$X^3-3X+1$

Do we first need to find its roots, then construct a splitting field $L$, then calculate $Gal(L/\mathbb{Q})$?
I am having difficulties finding roots. If we let the reduced cubic be: $U^2+qU+\frac{p^3}{27}=U^2+U+\frac{27}{27}=U^2+U+1$. The roots of this are: $x=\frac{-1 \pm \sqrt{-3}}{2}$
How do we use this to find the roots of the cubic?
Once I can decompose the polynomial I know that the Galois group will be $\{e\}, Z_2, A_3$ or $S_3$ depending on the degree of the splitting field and and how many linear factors there are,

$(X^3-2)(X^2+3)$

I have never encountered finding the Galois group of a degree $6$ polynomial but I am guessing that since it is factorised this eases things somewhat.
Let $f(X)=(X^3-2)(X^2+3)=(X-\sqrt[3]{2})(X^2+aX+b)(X-\sqrt{-3})(X+\sqrt{3})$
I am not sure how to find the coefficients of $X^2+aX+b$. Is it irreducible? 
Let $L$ be the splitting field of $f(X)$ over $\mathbb{Q}$ then (assuming $X^2+aX+b$ is irreducible) $L=\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$. 
If this is true what would $[\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3}), \mathbb{Q}]$ be?
I think this degree would be the order of the Galois group, so it could narrow down to one of $S_3, S_4, A_3, A_4...$ etc

REPLY [4 votes]: 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 $S_3$.