Cyclotomic polynomials and Galois groups According to this question I want to extend the question from there.
Lets consider again the galois extension $\mathbb Q(\zeta)/\mathbb Q$ where $\zeta$ is a primitive root of the $7^{th}$ cyclotomic polynomial.
I want to determine the minimal polynomial of $\zeta+\zeta^{-1}$ and $\zeta+\zeta^{2}+\zeta^{-3}$.
I know that one of the minimal polynomial has degree 2 and the other one degree 3, because $|Gal(L/K)|=6$. 
Well, started with squaring the first one, which yields $\zeta^2+2+\zeta^{-2}$, but how to continue?

The $7$th cyclotomic polynomial is 
$$\Phi_7(X)= X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$$
and so, after dividing by $X^3$ and rearranging terms we get:
$$\frac{\Phi_7(X)}{X^3} = X^3 + \frac{1}{X^3} + X^2 + \frac{1}{X^2} + X + \frac{1}{X} + 1
$$
The expression on the RHS can be expressed in terms of $X+\frac{1}{X}$. To do that, calculate:
$$(X+ \frac{1}{X})^3 = X^3 + \frac{1}{X^3} + 3( X + \frac{1}{X})\\
(X+\frac{1}{X})^2 = X^2 + \frac{1}{X^2} + 2$$
and therefore
$$\frac{\Phi_7(X)}{X^3} = (X+\frac{1}{X})^3 +(X+\frac{1}{X})^2 - 2(X + \frac{1}{X}) -1$$
or $$\frac{\Phi_7(X)}{X^3} = \Psi(\,X+\frac{1}{X})$$
where 
$$\Psi(Y) = Y^3 + Y^2 - 2 Y -1$$
Therefore, if $\lambda$ is a root of $\Phi_7$ then $\lambda+ \frac{1}{\lambda}$ is a root of $\Psi$. We also notice that $\Psi$ is irreducible over $\mathbb{Q}$. Indeed, the only potential rational roots are $\pm 1$ and these are not roots. Therefore, the minimal polynomial of $\lambda+ \frac{1}{\lambda}$ is $\Psi$.
Obs: The complex roots of $\Phi_7$ are $e^{\frac{2 k \pi i}{7}}$, for $1 \le k \le 6$. They group in $3$ pairs of inverse numbers $e^{\frac{2  \pi i}{7}}$ and $e^{\frac{2 \cdot 6 \pi i}{7}}$, $e^{\frac{2\cdot 2  \pi i}{7}}$ and $e^{\frac{2\cdot 5 \cdot 6 \pi i}{7}}$, $e^{\frac{2 \cdot 3 \pi i}{7}}$ and $e^{\frac{2\cdot 4 \cdot 6 \pi i}{7}}$. We conclude that the roots of the polynomial $\Psi$ are 
$2\cos (\frac{2 \pi}{7}), 2\cos (\frac{4 \pi}{7}), 2\cos (\frac{6 \pi}{7})$
Notice the equality 
$$\{2, 4, 1\} = \{x \in (\mathbb{Z}/7)^{\times}\ | x \ \text{is a square}\}$$
Therefore, for evey $a \in (\mathbb{Z}/7)^{\times}$ we have
\begin{eqnarray}
a \cdot \{2, 4, 1\} &=&\{2, 4, 1\}  \text{ if } a \ \text{is a square} \\
a \cdot \{2, 4, 1\} &=& \{3,5,6\}  \text{ if } a \ \text{is not a square} 
\end{eqnarray}
We conclude that the automorphism $\rho_a$ of $\mathbb{Q}(\zeta)$, $\zeta\mapsto \zeta^a$ invariates each of $\zeta + \zeta^2 + \zeta^ 4$, $\zeta ^3 + \zeta^5 + \zeta^6$ or switches them. Hence these elements are conjugate. Let's find their sum and product.
$$(\zeta + \zeta^2 + \zeta^ 4) +( \zeta ^3 + \zeta^5 + \zeta^6)=-1 \\
 (\zeta + \zeta^2 + \zeta^ 4)  (  \zeta^3 + \zeta^5 + \zeta^6) = \zeta^4+\zeta^6+1+\zeta^5+1+\zeta+ 1+\zeta^2+\zeta^3=\\
2 + (1 + \zeta + \zeta^2+\zeta^3+\zeta^4+\zeta^5+\zeta^6)=2
$$
Therefore $\zeta + \zeta^2 + \zeta^ 4$ and $\zeta^3 + \zeta^5 + \zeta^6$ are the roots of 
$$Z^2 + Z + 2$$
Obs: 
For general $n$ and $\zeta$ a primitive root of unity of order $n$, and t $H$ is a subgroup of $(\mathbb{Z}/n)^{\times}$, consider the sum
$$\sum_{g \in H} \zeta^{g}$$
Then its conjugates will be 
$$\sum_{g \in aH} \zeta^{g}$$
where $aH$ are the cosets of $H$ in $ (\mathbb{Z}/n)^{\times}$  $\tiny{\text{(Gauss sums)}}$
