Let $a \in \mathbb{C}$, $Y=X^3+3X^2+aX \in \mathbb{C}(X)$. When $\mathbb{C}(X)/ \mathbb{C}(Y) $is Galois extension?
I know that $\mathbb{C}(X)$ is perfect, so it is sufficient to consider when is normal expansion. But, I can't do it because I don't know basis.
Let us fix an $a\in\Bbb C$. Assume that the extension $\Bbb C(X)$ over $\Bbb C(Y)$ is Galois. Let $P$ be the polynomial $$ P(T)=T^3+3T^2+aT-Y\in\Bbb C(Y)[T] $$ over the ground field of the extension. One root of it is $X$, since plugging in $X$ instead of $T$ delivers $P(X) =0$. The polynomial is thus divisible by $(T-X)$, and the quotient is the $Q(T)$ from $$ P(T)=(T-X)\cdot\underbrace{(T^2+(X+3)T+(X^2 +3X+a)}_{\text{Notation: }Q(T)}\ . $$ This polynomial has as roots conjugates of $X$, one root of $P$, so it splits over $\Bbb C(Y)$.
The roots of $Q(T)$ in $\Bbb C(X)$ are of the shape $\frac 12\Big(\ -(X+3)\pm\sqrt \Delta\ \Big)$, where $\Delta$ is the usual discriminant, here built w.r.t. $T$, $$ \begin{aligned} \Delta&=(X+3)^2-4 (X^2 +3X+a) \\ &= -3X^2-6X-(4a-9)\ . \end{aligned} $$ The condition is now to get a rational function for $\sqrt \Delta$. (Else we need to pass to $\Bbb C(X)[\sqrt\Delta]$, a true quadratic extension, to have a split.)
Note that $-3X^2-6X-3=-3(X+2)^2$ is already a square in $\Bbb C(X)$. We obtain one solution, $$a=3\ .$$ Using the arithmetic in the factorial ring $\Bbb C[X]$ we see that there is no other solution, since an equation of the form $\Delta=R^2$ in $\Bbb C(X)$ implies that $\Delta$ is a square of a polynomial of degree one.