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 $\mathbb{Q}$.

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.

We use a fundamental theorem of Galois theory, that the degree of a Galois extension equals the order of the Galois group of that extension. Note that extensions obtained by adding roots of a polynomial with coefficients in the field are automatically Galois extensions.

The logic is that since $f(x) \in \mathbb{Q}[x]$ is a cubic, its Galois group (i.e. the Galois group of a splitting field) will be a subgroup of $S_3$ which has order $6$.

More explicitly, let $x_1, x_2, x_3$ be the (complex) roots of $f$. Then certainly $K=\mathbb{Q}(x_1, x_2, x_3)$ is a splitting field. The Galois group $G$ is the set of those automorphisms of $K$ that fix $\mathbb{Q}$, and so are determined by how they act on the roots. However, since any automorphism fixes $f$, the image of a root under any automorphism is still a root, so $G$ permutes the roots and hence $G$ is a subgroup of $S_3$.

Now the second part is actually finding polynomials that have Galois groups $1$, $C_2$, $C_3 = A_3$ and $S_3$.

$1$ is easy enough: just take the product of three linear polynomials such as $(x-1)(x-2)(x-3)$.

For $C_2$, you need a quadratic polynomial with non-rational roots, for instance $(x-1)(x^2+1)$.

For $S_3$, you can repeat the idea in $C_2$ but this time giving a non-rational root to the linear part, e.g. $x^3 -2$.

Getting a polynomial with $C_3$ is perhaps the hardest one, but with a bit of trial-and-error or some additional insight on an object called "the discriminant" $x^3 -3x+1$ is an example.