Degree of the minimal polynomial of $\sum\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$. The polynomial $$f(x)=
x^8-8x^6+20x^4-16x^2+2=((x^2-2)^2-2)^2-2=0$$ has $$y=\sqrt{2+\sqrt{2+\sqrt{2}}}$$ one of its roots.
How do I determine the degree of its spliting field and how do I determine its Galois Group?
What I know is the other roots are
$$\pm
\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$$ and these $8$ elements of $\mathbb{R}$ are the only roots of this polynomial. But I am not sure how do I proceed further.
I am computing this because I was trying to find the degree of the minimal polynomial of the element
$$ \sum \sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$$ and this element is contained in the splitting field of $f(x)$ and these are fixed by conjugation(inaccurate). I need to work a bit more on this. I would appreciate if I can get an idea about it galois groups.
 A: Preamble:
Let $\zeta$ be a primitive $32$nd root of unity.
First, note that $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\cong(\mathbb{Z}/32\mathbb{Z})^\times\cong C_2\times C_8$. In fact, every element in $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ can be writen as $\sigma^i\tau^j$ for some unique $i\in\mathbb{Z}_8,j\in\mathbb{Z}_2$, where $\sigma:\zeta\rightarrow\zeta^3$ and $\tau:\zeta\rightarrow\zeta^{-1}$ is the complex conjugate. Furthermore, note that $\mathbb{Q}(\cos(\pi/16))$ is a subfield of $\mathbb{Q}(\zeta)$ that is invariant under the automorphism subgroup $\{\text{id},\tau\}$; therefore by the Fundemental Theorem of Galois Theory, $\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})/\{\text{id},\tau\}\cong C_8$.
Derivation:
Now, let $f(x)$ be your polynomial, and let $E$ be the splitting field of $f(x)$. Note that (as was pointed out in the comments)
\begin{equation}
\sqrt{2+\sqrt{2+\sqrt{2}}}=2\cos\left(\frac{\pi}{16}\right)
\end{equation}
This means that $E/\mathbb{Q}(\cos(\pi/16))$ is a possibly trivial field extension. In fact, we have that
\begin{equation}
8=|\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})|=[\mathbb{Q}(\cos(\pi/16)):\mathbb{Q}]\leq[E:\mathbb{Q}]\leq\deg(f)=8
\end{equation}
so $[\mathbb{Q}(\cos(\pi/16)):\mathbb{Q}]=[E:\mathbb{Q}]$ which implies that $E=\mathbb{Q}(\cos(\pi/16))$. As discussed before, this means that $\text{Gal}(E/\mathbb{Q})\cong C_8$, and therefore the degree of the splitting field of $f(x)$ is $8$.
Addendum:
Let $\alpha=\sum\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$ be your sum. There are multiple ways to go about calculating $[\mathbb{Q}(\alpha):\mathbb{Q}]$ which is the degree of the minimal polynomial of $\alpha$.
One way is to notice that
\begin{equation}
\alpha=2\left[\cos\left(\frac{\pi}{16}\right)+\cos\left(\frac{3\pi}{16}\right)+\cos\left(\frac{5\pi}{16}\right)+\cos\left(\frac{7\pi}{16}\right)
\right]
\end{equation}
which you can get by noting that $2\cos(k\pi/{16})$ for $k\in\{1,3,5,7\}$ are precisely the positive numbers which are the images of $2\cos(\pi/16)$ under the automorphisms in $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$, meaning they must be some permutation of the numbers $\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$ which are the positive roots of $f(x)$. Now, we can note that $\alpha$ is only fixed by id and $\tau$, which by the same logic as in the above proof, tells us that $[\mathbb{Q}(\alpha):\mathbb{Q}]=8$.
Alternatively, one can further simplify to see that $\frac{1}{\alpha}=\cos\left(\frac{7\pi}{16}\right)$, which means that $\mathbb{Q}(\cos(\pi/16))=\mathbb{Q}(\cos(7\pi/16))\subseteq\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\cos(\pi/16))$ which again confirms that $[\mathbb{Q}(\alpha):\mathbb{Q}]=8$.
