Find Galois group of $x^3+x^2-2x-1\in\mathbb{Q}[x]$ 
Find Galois group of $f(x)=x^3+x^2-2x-1\in\mathbb{Q}[x]$ over $\mathbb{Q}$.

This is not a duplication of this question because in the other question they are given a further data.
My attempt:
Let $E$ be the splitting field of $f$ over $\mathbb{Q}$. Let $\zeta=\exp(\frac{2\pi i}{7}), \alpha=\zeta+\zeta^{-1}$. We can notice that $f(\alpha)=0$, $f$ is irreducible in $\mathbb{Q}$ and $\deg(f)=3$. Thus $\text{irr}(\alpha,\mathbb{Q})=f$. Furthermore, we know that $[\mathbb{Q}(\zeta):\mathbb{Q}]=6$. We know that $\alpha\in E$. Thus we have this fields building:

And here I stuck. Thanks for helping.
 A: This is a pedestrian answer first, then we restate in the framework of Galois theory.

I will denote by $u$ a fixed primitive $7$.th root of unity (instead of $\zeta$ or even $\zeta_7$, which is hard to type.)
Let $a$ be the element $u+u^{-1}$. We know its conjugates, and introduce letters for them:
$$
\begin{aligned}
a &= u^1 + u^{-1} = u^1 + u^6\ ,\\
b &= u^2 + u^{-2} = u^2 + u^5\ ,\\
c &= u^3 + u^{-3} = u^3 + u^4\ .
\end{aligned}
$$
Then the given polynomial decomposes as
$$
X^3 + X^2 - 2X - 1
=(X-a)(X-b)(X-c)\ .
$$
To see this, we check the relations of Vieta.
$$
\begin{aligned}
a+b+c 
&= 
u+u^2+u^3+u^4+u^5+u^6
\\
&=
-1\ ,\\[2mm]
ab+ac+bc
&=ab + c(a+b)
\\
&=u^3(1+u^5)(1+u^3) + u^3(1+u)(-1-u^3-u^4)\\
&=u^3(1+u^3+u^5+u)-u^3(1+u^3+u^4+u+u^4+u^5)\\
&=-u^3(u^4+u^4)\\
&=-2\ ,\\[2mm]
abc
&=u^{1+2+3}(1+u^5)(1+u^3)(1+u)\\
&=u^6(1+u+u^3+u^5)(1+u)\\
&=u^6(1+u+u^3+u^5+u+u^2+u^4+u^6)\\
&=u^6\cdot u\\
&=1\ .
\end{aligned}
$$
The computation was done explicitly to show with bare hands that $a,b,c$ are the conjugates of $a$ in $E$, which is thus $\Bbb Q(a)$. (We have $b=a^2-2$ and $c=a^3-3a$.)

Let us give the Galois touch to the situation. The Galois group of the cyclotomic extension $\Bbb Q(u)$ over $\Bbb Q$ has the Galois substitutions $s_k:u\to u^k$ for $k\in\{1,2,3,4,5,6\}= \Bbb F_7^\times$, and the structure corresponds to the multiplicative structure of $\Bbb F_7^\times$, because $s_k\circ s_n\ u=s_k(s_n(u))=s_k(u^n)=(u^n)^k=u^{kn}=s_{kn}\ u$.
In particular, $s_{-1}=s_6$ is an element of order two, which stabilizes $a=u+u^{-1}$. There is a correspondence now between the subgroups of $\operatorname{Gal}(\Bbb Q(u):\Bbb Q)\cong \Bbb F_7^\times\cong(\Bbb Z/6,+)$ and the intermediate fields, normal corresponding to normal. But the Galois Group is commutative, each subgroup is normal. The subgroup of order two $\{1,s_6\}$ corresponds to $\Bbb Q(a)$. The Galois group of $\Bbb Q(a)$ over $\Bbb Q$ is $\cong \Bbb Z/3$, seen as the group of all six Galois substitutions of $\Bbb Q(u)$, modulo the subgroup generated by $s_6$, since $s_6$ acts trivially on $a$. 
