<s>[INST] Find rational numbers $c_0$, $c_1$, ..., $c_{d-1}$ such that $2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$. Let $a = \sqrt{2 - \sqrt{2}}$.
(1) Find the minimal polynomial of $a$ over $\mathbb{Q}$.
(2) Let $d = [\mathbb{Q}(a) ~:~ \mathbb{Q}]$. Compute $d$ and find rational numbers $c_0$, $c_1$, ...,
$c_{d-1}$ such that
$\hspace{10mm} 2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$.
(3) Prove or disprove that all roots of the minimal polynomial of $a$ over $\mathbb{Q}$ belongs to $\mathbb{Q}(a)$.
$\textbf{My Attempt:}$
For question (1):
Since, $a = \sqrt{2 - \sqrt{2}} \implies (2 - a^2)^2 = 2 \implies a^4 - 4a^2 + 2 = 0$.
Notice that $x^4 - 4 x^2 + 2$ is irreducible over $\mathbb{Q}$ and $a^4 - 4a^2 + 2 = 0$ and $x^4 - 4 x^2 + 2$ is monic polynomial.
So, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$.
For question (2):
Since, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$. Then, the degree of it is $4$.
So, $d = [\mathbb{Q}(a) ~:~ \mathbb{Q}] = $ the degree of the minimal polynomial of $a$ over $\mathbb{Q}$ which equals to $4$.
For question (3):
Since, the minimal polynomial of $a$ over $\mathbb{Q}$ is $x^4 - 4x^2 + 2$.
Which $x^4 - 4x^2 + 2 = (x - \sqrt{2 - \sqrt{2}})(x + \sqrt{2 - \sqrt{2}})(x - \sqrt{2 + \sqrt{2}})(x + \sqrt{2 + \sqrt{2}})$.
Then, $\pm \sqrt{2 - \sqrt{2}}$, $\pm \sqrt{2 + \sqrt{2}}$ are all roots to $x^4 - 4x^2 + 2$.
But $-\sqrt{2 + \sqrt{2}}$ and $\sqrt{2 + \sqrt{2}}$ are not belongs to $\mathbb{Q}(a)$.
So, we disproved that all roots of the minimal polynomial of $a$ over $\mathbb{Q}$ belongs to $\mathbb{Q}(a)$.
$\textbf{My Question:}$
First, for question (2) I am stuck on finding the rational number $c_0$, $c_1$, ...,
$c_{d-1}$ such that
$\hspace{10mm} 2 a^{-1} - 3 a^6 = c_0 + c_1 a + \cdots + c_{d-1} a^{d-1}$.
I can only find out that $2a^{-1} - 3a^6 = (2 + \sqrt{2}) a - (18 - 12\sqrt{2}) a$
which I can not further convert to $c_0 + c_1 a + c_2 a^2 + c_3 a^3$ where $c_0, ..., c_3 \in \mathbb{Q}$.
Also, are the other question correct ?
 [/INST] Note that
$$
a^4 - 4a^2 + 2 = 0 \implies  2 = -a^4 + 4a^2 \implies 2a^{-1} = -a^3+4a \\
a^4 - 4a^2 + 2 = 0 \implies a^4 = 4a^2-2 \implies a^6 = 4a^4-2a^2 = 4(4a^2-2)-2a^2 
$$
and, since $a^2 = 2-\sqrt2$,
$$
\begin{align}
\pm \sqrt{2+\sqrt2} &= \pm \sqrt{\frac{(2+\sqrt2)(2-\sqrt2)}{2-\sqrt2}} \\ &= \pm \sqrt{\frac{2}{2-\sqrt2}}
\\ &= \pm \frac{\sqrt2}{\sqrt{2-\sqrt2}} \\ &= \pm \frac{2-a^2}{a} \in \mathbb Q(a).
\end{align}$$
</s>