Your algebra seems to check out.
I would have simplified the working of finding the minimal polynomial of $\alpha$ like so:
$\alpha^3 = \left(\zeta + \dfrac{1}{\zeta}\right)^3 = \zeta^3 + 3\zeta + \dfrac{3}{\zeta} + \dfrac{1}{\zeta^3}$
$= \zeta^3 + \dfrac{1}{\zeta^3} + 3\alpha = \zeta^3 + \zeta^4 + 3\alpha$
and:
$\alpha^2 = \left(\zeta + \dfrac{1}{\zeta}\right)^2 = \zeta^2 + 2 + \dfrac{1}{\zeta^2} = \zeta^2 + \zeta^5 + 2$
so that:
$\alpha^3 + \alpha^2 = \zeta^3 + \zeta^4 + \zeta^2 + \zeta^5 + 3\alpha + 2$
$= \zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 + \zeta^6 + 2\alpha + 2$
$= -1 + 2\alpha + 2$, so that:
$\alpha^3 + \alpha^2 - 2\alpha - 1 = 0$.
However, this only shows that $\alpha$ is a root of the cubic $p(x) = x^3 + x^2 - 2x - 1$. To show this is the *minimal* polynomial you have two options:
1) Show it is irreducible over $\Bbb Q[x]$. I would use the rational root theorem (Gauss' lemma).
2) Show that $\sigma_6$ generates a subgroup of order 2 of $\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)$, and that the fixed field of this subgroup is $\Bbb Q(\alpha)$.
You also claim without proof that the roots of $p$ are:
1) $\alpha = \zeta + \zeta^6$ (no need to prove this)
2) $\zeta^2 + \zeta^5$
3) $\zeta^3 + \zeta^4$
I would mention that this is because any automorphism of $\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)$ sends a root of $p$ to another root of $p$.
Since $\Bbb Q(\alpha)$ contains all the roots of $p$, it's a normal extension (thus Galois). Clearly, its Galois group is:
$\dfrac{\text{Gal}(\Bbb Q(\zeta)/\Bbb Q)}{\text{Gal}(\Bbb Q(\zeta)/\Bbb Q(\alpha))} \cong \Bbb Z_6/\langle 3\rangle \cong \Bbb Z_3$
(recall that $3$ is the sole element of order 2 in $\Bbb Z_6$-just as complex-conjugation restricted to $\Bbb Q(\zeta)$ (the map $\sigma_6 : \zeta \mapsto \zeta^6 = \zeta^{-1} = \overline{\zeta}$) is the sole automorphism of the Galois group of $\Phi_7(x)$ of order 2).
