Let $f(x)= x^3+ax^2+bx+c \in \mathbb{Q}[x]$. Show that the splitting field of $f$ over $\mathbb{Q}$ has degree 1, 2, 3 or 6 over $\mathbb{Q}$. QUESTION: Let $f(x)= x^3+ax^2+bx+c \in \mathbb{Q}[x]$. Show that the splitting field of $f$ over $\mathbb{Q}$ has degree 1, 2, 3 or 6 over $\mathbb{Q}$.
The professor gave us this hint, but I still not understanding. I need to solve this in a step by step way. Using his tips.
HINT: The greatest difficulty would be to show that it cannot be greater than 6. Then, is enough to choose some values for $a, b$ and $c$. Try to find on the part of Galois that the extension has degree $\leq n!$. You need to find polynomials in that way that have splitting fields of degrees $1, 2, 3$ and $6$. And then show that it can't be bigger than that. It cannot be greater than 6 because this happens in the worst case ... It has a real root that has a degree $\leq3$ (it always exists since the polynomial has an odd degree, using the intermediate value theorem) and a complex one (which can also be real) of degree $\leq 2$. Then the degree of extension $\leq 6$. We use intermediate value theorem because polynomials of odd degree have a real root.
I really appreciate your help if you would take a time to help me.

Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Since $\mathbb{Q}$ has characteristic zero, the extension is separable, and it is a splitting field so it is normal. Therefore $L/\mathbb{Q}$ is a Galois extension.
We know that the Galois group $G = \operatorname{Gal}(L/\mathbb{Q})$ acts faithfully on the roots of $f$ in $L$. There are three such roots $\alpha_1,\alpha_2, \alpha_3$ say, so $G$ may be viewed as a group of permutations of $\{\alpha_1,\alpha_2,\alpha_3\}$, which makes it a subgroup of the symmetric group $S_3$. Since $S_3$ has order $6$, it follows that the order of $G$ divides $6$, so it is $1,2,3$ or $6$.
It is a standard result of Galois theory that the degree of a Galois extension equals the order of its Galois group, so $[L : \mathbb{Q}] = \lvert G \rvert$ is $1, 2, 3$ or $6$.
Finally, Piquito's comment shows that each of these possibilities does actually occur.
