Question about a Galois group I am working on a problem which seems to be troubling me quite a lot. This is how it goes:
Let $\alpha=\sqrt{\frac{2\sqrt{3}-3}{3}}$ and $\beta=\sqrt{\frac{2\alpha}{\sqrt{3}}}$. It is easy to show that the minimal polynomial of $\alpha$ and $\beta$ are $3x^4+6x^2-1$ and $27x^8+72x^4-16$, respectively. One can see that $\mathbb{Q}(\beta,i)$ contains "the" splitting field of $3x^4+6x^2-1$. The part I am having trouble with is to show that this field also contains "the" splitting field of $27x^8+72x^4-16$.
I was wondering if anyone could point me in the right direction.
 A: It may be of help to look at the roots of $27x^8 +72x^4 -16$. In order to compute the roots you can think of it as a quadratic polynomial via a change of variables $y = x^4$.
A: Let's check the roots of the equation $27x^8+72x^4-16=0$. Putting $t=x^4$ we get the equation $27t^2+72t-16=0$ with roots $$t=\frac{-12\pm8\sqrt{3}}{9}=\beta^4,-\frac{16}{27\beta^4}$$ Hence the roots of original equation are given by $$x=\pm\beta, \pm i\beta, \pm\frac{1\pm i}{\beta}\sqrt[4]{\frac{4}{27}}$$ so our job is done if we can show that $\sqrt[4]{4/27}\in\mathbb{Q}(\beta,i)$.
We can in fact show that $\sqrt[4]{4/27}\in \mathbb{Q} (\beta) $ or equivalently $a=\sqrt[4]{12}\in\mathbb{Q}(\beta)$. We note that $$a^2=2\sqrt{3}=\frac{9\beta^4+12}{4}$$ or
\begin{align}
9\beta^4&=4a^2-12\notag\\
&=a^2(4-a^2)\notag\\
&=a^2(4-2\sqrt{3})\notag\\
&=a^2(\sqrt{3}-1)^2\notag
\end{align}
and then $$3\beta^2=a(\sqrt{3}-1)$$ Note that $\sqrt{3}=(9\beta^4+12)/8\in\mathbb{Q}(\beta)$ and hence by above equation $a\in\mathbb{Q} (\beta) $.

While typing my answer I remembered a previous answer of mine which contained similar calculation. Perhaps the equation there is related to the one given here via some simple substitution.
