Galois Group of $x^{14}+x^7-1$ over $\mathbb{Q}$ So consider the polynomial $f(x)=x^{14}+x^7-1$ defined over $\mathbb{Q}$.
We want to determine its Galois Group.
So let's look for the splitting field, $L$ say, to give us an idea of the size of the Galois Group.
Note that letting $y=x^7$ we see that $y$ satisfies $y^2+y-1=0$ gives us that $y=\frac{1\pm \sqrt5}{2}$ as roots.
Clearly these must be in the splitting field of $f$ since for any root of $f$ each of its powers must be in the splitting field and these are the seventh powers of the roots of $f$.
So $\mathbb{Q}(\sqrt5) \subset L$ and thus $2| [L:\mathbb{Q}]$.
Now I don't  know where to go from here.
We see that $\alpha^7=\frac{1\pm \sqrt5}{2}$ for any root $\alpha$, but how do I work out the size of the extension $[L:\mathbb{Q}(\sqrt5)]$? 
 A: Extending my comment to a partial solution showing that $[L:\Bbb{Q}]=84$.
Let $\alpha$ be the real number with the property $\alpha^7=(1+\sqrt 5)/2$, and
let $\zeta=e^{2\pi i/7}$ be the given seventh primitive root of unity. As $(-1\pm\sqrt5)/2$ are the zeros of the polynomial $y^2+y-1$, and they are negative reciprocals of each other, we see that the zeros of $f(x)=x^{14}+x^7-1$ are $-\alpha\zeta^k, \alpha^{-1}\zeta^k$, for
$k=0,1,\ldots,6$.
This implies that $L=\Bbb{Q}(\sqrt5,\alpha,\zeta)$. Let $K=\Bbb{Q}(\sqrt5)$.
I first claim that $[K(\zeta):K]=6$. As $7$ is a prime, it is well known that $[\Bbb{Q}(\zeta):\Bbb{Q}]=6$. Adjoining $\sqrt5$ won't change the degree of the extesnion. For if we had $[K(\zeta):\Bbb{Q}]=6$, then this would imply that $K(\zeta)=\Bbb{Q}(\zeta)$, in other words we would have $\sqrt5\in \Bbb{Q}(\zeta)$. But $\Bbb{Q}(\zeta)$ is a cyclic extension of degree six over $\Bbb{Q}$, so it has a unique quadratic subfield. It is also known (Gauss' sum) that this quadratic extension is $\Bbb{Q}(\sqrt{-7})$. This implies that
$\sqrt5\notin \Bbb{Q}(\zeta)$ proving the first claim.
The second claim is that $[K(\alpha):K]=7$. In other words I claim that the polynomial
$$
p(x)=x^7-\left(\frac{1+\sqrt5}2\right)\in K[x]
$$
is irreducible. This needs a bit of algebraic number theory and basics of finite fields.
The ring of integers of $K$ is ${\mathcal O}=\Bbb{Z}[\frac{1+\sqrt5}2].$ This is a principal ideal domain, so it suffices to show that $p(x)$ is irreducible in ${\mathcal O}[x]$. Consider the prime ideal $\mathfrak{p}=\langle7+2\sqrt5\rangle\subset{\mathcal O}$.
We easily see that ${\mathcal O}/\mathfrak{p}\cong\Bbb{Z}_{29}$ (the prime $29$ splits
in $K$, and $\mathfrak{p}$ is one of the ideals of $\mathcal O$ lying above $29$).
Because $11\equiv\sqrt5\pmod{\mathfrak{p}}$, we see that the projection of $p(x)$ in
${\mathcal O}/\mathfrak{p}[x]\cong\Bbb{Z}_{29}[x]$ is
$$
\overline{p}[x]=x^7-\overline{6}\in\Bbb{Z}_{29}[x].
$$
The residues class $\overline{6}$ is of order $14$ modulo $29$ - a multiple of seven!! This implies that any zero of $\overline{p}[x]$ (in some extension of $\Bbb{Z}_{29}$) is of order $49$ or $98$. The lowest exponent $k>0$ such that $29^k\equiv1\pmod{49}$ is $k=7$. So the roots of $\overline{p}[x]$ belong to the field $\Bbb{F}_{29^7}$. Thus $\overline{p}(x)$
is irreducible and, consequently so is $p(x)$.
As six and seven are coprime together the two claims imply that $[L:K]=42$, and thus
$[L:\Bbb{Q}]=84$. Have fun figuring out the Galois group as a subgroup $S_{14}$!
