Complex roots of minimal polynomial using Galois

Question

For the algebraic number $-3\sqrt6 + 3\sqrt[3]5$ I found the minimal polynomial over $\mathbb{Q}$ (degree = 6). So now I need to find complex roots of this polynomial, but I don't know how.
I have an idea: I need to look at automorphisms (elements of Galois group) of this polynomial, and observe that it isomorphic to some symmetric group etc. But still I don't know how to find these roots exactly.

Answer 1

The other roots are the numbers of the form $\pm3\sqrt6+\left(-\frac12+\frac{\sqrt3}2i\right)^k3\sqrt[3]5$, with $k\in\{0,1,2\}$.

Answer 2

Begin by seeing that $$\mathbb{Q}(-3\sqrt{6}+3\sqrt[3]{5})=\mathbb{Q}(\sqrt{6},\sqrt[3]{5}).$$

Now, this is not a normal field, since the minimal polynomial for $\sqrt[3]{5}$ does not split. The normal closure of this would then need to be

$$\mathbb{Q}(\sqrt{6},\sqrt[3]{5},\omega)$$

where $\omega$ is a cube root of unity.

This is an extension of degree $12$ over $\mathbb{Q}$, and so we have $12$ automorphisms that fix $\mathbb{Q}$. We can see that these automorphisms are just the combinations of automorphisms moving around the roots of $X^3-1$, $X^3-5$, and $X^2-6$.

In particular, $\sqrt{6}$ can be mapped to $\sqrt{6}$ or to $-\sqrt{6}$.

$\sqrt[3]{5}$ can be mapped to $\omega\sqrt[3]{5}$, $\omega^2 \sqrt[3]{5}$ or $\sqrt[3]{5}$.

Since the Galois Group acts transitively on the roots of the minimal polynomial, this gives us all of the possibilities.