If $c$ is a complex root of a cubic $a(x)\in\mathbb{Q}[x]$, show that $\mathbb{Q}(c)$ is the splitting field of $a(x)$ over $\mathbb{Q}$.
For this, we must show that $\mathbb{Q}(c)$ contains all three roots of $a(x)$. It certainly contains $c$.
There are two other roots of $a(x)$, one of which is real, and the other of which is the conjugate of $c$. Suppose $c=p+qi$, then $\overline{c}=p-qi=\dfrac{p^2+q^2}{p+qi}$, so $\overline{c}\in\mathbb{Q}(c)$. (Hmm... actually I just realized $p,q$ might not belong to $\mathbb{Q}$.. maybe trouble here)
What about the last root? We know that the sum of the three roots is a rational number.
This is false, and in fact nearly the opposite is true: $\mathbb{Q}(c)$ is never the splitting field of $a(x)$ if $a(x)$ is irreducible. Indeed, if $\mathbb{Q}(c)$ were the splitting field, then $\mathbb{Q}(d)$ would also be the splitting field for any other root $d$ of $a(x)$ (since $a(x)$ is irreducible so $\mathbb{Q}(c)\cong\mathbb{Q}[x]/(a(x))\cong\mathbb{Q}(d)$). But if $d$ is the real root of $a(x)$, then $\mathbb{Q}(d)\subset\mathbb{R}$ so it cannot contain all the roots of $a(x)$.