Let $f(x) = x^4 + 2x^2 - 6$ and let E be the splitting field of $f(x)$ over $\mathbb{Q}$.
(a) Find roots $\alpha$ and $\beta$ of $f(x)$ such that $E = \mathbb{Q}(\alpha, \beta)$. What are the other roots of $f(x)$ ?
(b) Compute $[\mathbb{Q}((\alpha + \beta)^2) : \mathbb{Q}]$.
(c) Compute $[\mathbb{Q}(\alpha + \beta) : \mathbb{Q}]$.
$\textbf{My Attempt:}$
(a) Notice that $f(x) = x^4 + 2x^2 - 6 = (x-\sqrt{-1+\sqrt{7}})(x+\sqrt{-1+\sqrt{7}})(x-\sqrt{-1-\sqrt{7}})(x+\sqrt{-1-\sqrt{7}})$.
Then, Case 1: $\alpha = \sqrt{-1+\sqrt{7}}$ and $\beta = \sqrt{-1-\sqrt{7}}$;
or Case 2: $\alpha = -\sqrt{-1+\sqrt{7}}$ and $\beta = -\sqrt{-1-\sqrt{7}}$.
Which in both of the cases $\alpha$ cannot be represent in terms of $\beta$ and $\beta$ cannot be represent in terms of $\alpha$.
So, $E = \mathbb{Q}(\sqrt{-1+\sqrt{7}}, \sqrt{-1-\sqrt{7}}) = \mathbb{Q}(-\sqrt{-1+\sqrt{7}}, -\sqrt{-1-\sqrt{7}})$. $\textbf{But how do I prove this is true ?}$
(b) for case 1: if $\alpha = \sqrt{-1+\sqrt{7}}$ and $\beta = \sqrt{-1-\sqrt{7}}$
Then, $(\alpha + \beta)^2 = (\sqrt{-1+\sqrt{7}} + \sqrt{-1-\sqrt{7}})^2 = -2 + 2i\sqrt{6}$.
Notice that $-2$ and $2$ are in $\mathbb{Q}$. So, $\mathbb{Q}((\alpha + \beta)^2) = \mathbb{Q}(-2 + 2i\sqrt{6}) = \mathbb{Q}(i\sqrt{6})$.
Let $c = i\sqrt{6} \implies c^2 = -6 \implies c^2 + 6 = 0$.
Also, by Eisenstein’s criterion we know that $x^2 + 6$ irreducible over $\mathbb{Q}$. And $x^2 + 6$ is monic polynomial.
So, the minimal polynomial of $c$ over $\mathbb{Q}$ is $m_{c, \mathbb{Q}}(x) = x^2 + 6$.
Hence, $[\mathbb{Q}((\alpha + \beta)^2) : \mathbb{Q}] = [\mathbb{Q}(i\sqrt{6}) : \mathbb{Q}] = $ degree of $m_{c, \mathbb{Q}}(x)$ which is equal to $2$.
(c) Since, $\mathbb{Q}(\alpha + \beta) = \mathbb{Q}(\alpha, \beta)$. $\textbf{But how do I prove this is true ?}$
Then, $[\mathbb{Q}(\alpha + \beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\alpha)] \times [\mathbb{Q}(\alpha) : \mathbb{Q}]$
$= [\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\beta)] \times [\mathbb{Q}(\beta) : \mathbb{Q}] = 4 \times 4 = 8$.
By Tower Law.
$\textbf{Are these correct ?}$
$\textbf{Also, how do I prove that}$ $E = \mathbb{Q}(\sqrt{-1+\sqrt{7}}, \sqrt{-1-\sqrt{7}}) = \mathbb{Q}(-\sqrt{-1+\sqrt{7}}, -\sqrt{-1-\sqrt{7}})$
$\textbf{and}$ $\mathbb{Q}(\alpha + \beta) = \mathbb{Q}(\alpha, \beta)$ $\textbf{are true ?}$
Let me abbreviate your four roots to $r,s,t,u$. Note $s=-r$ and $u=-t$, so any field containing $r$ automatically contains $s$, and any field containing $t$ automatically contains $u$. So, extending the rationals by $r$ and $t$ gives you all four of the roots and hence gives you the splitting field $E$.
Now, let $x=r+t$. You have shown $x^2=-2+2i\sqrt6$. So, $(x^2+2)^2=-24$, so $x^4+4x^2+28=0$. If you can show that this polynomial is irreducible over the rationals, then you have proved that $x$ has degree four over the rationals, which answers (c).