Use Galois Theory to prove the existence of $A$ and $B$ such that $\mathbb{Q}(\sqrt{6+3\sqrt{3}})=\mathbb{Q}(\sqrt{A}, \sqrt{B})$

Question

Use Galois Theory to prove the existence of $A$ and $B$ such that $\mathbb{Q}(\sqrt{6+3\sqrt{3}})=\mathbb{Q}(\sqrt{A}, \sqrt{B})$

So $\mathbb{Q}(\sqrt{6+3\sqrt{3}})$ is the field of rational numbers with the extra element generated by $\sqrt{6+3\sqrt{3}}$. Is this correct? How could we proceed using Galois Theory?

Find these $A$ and $B$

From inspection I think, without loss of generality, $A=\sqrt{2}$ and $B=\sqrt{3}$.

How could I prove this?

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Edit, linked question gives useful approach but further answers are welcome (and useful)

Answer 1

Notice your element satisfies

$$(x^2-6)^2-27=0\iff x^4-12x^2+9=0$$

Then if you adjoin the square root of $3$ you get it to split as

$$x^2-6=3\sqrt3$$

i.e. the degree of the total extension is $4$, so the Galois group has order $4$. But there are only two groups of order $4$, and since you can exhibit more than one element of order $2$, you know the Galois group is the Klein $4$ group. But then all such extensions with that Galois group are of the form $\Bbb Q(\sqrt{A},\sqrt{B})$, completing the proof.