Finding the Galois group of $x^6-x^5+x^4-x^3+x^2-x+1$ over $\Bbb{Q}$

I was trying to find the Galois group of $x^6-x^5+x^4-x^3+x^2-x+1$ over $\Bbb{Q}$, but I'm confused, because I want to find the splitting field for this polynomial... but I have to find the roots first and I'm not sure how to (since the polynomial is of degree 6). So I was wondering if anybody could help me with that...

Perhaps you could note that $\alpha$ is a root of $$f = x^6-x^5+x^4-x^3+x^2-x+1,$$ if and only if $-\alpha$ is a root of $$g = x^6+x^5+x^4+x^3+x^2+x+1 = \frac{x^7 - 1}{x - 1}.$$ Clearly, the splitting fields of $f$ and $g$ are the same, and you should have seen something about cyclotomic fields...
Alternatively, as noted by @Kunnysan, a root $\alpha$ of $f$ is a $14$-th root of unity (as $\alpha^{14} = (\alpha^{7})^{2} = (-1)^{2} = 1$), and the $14$-th roots of unity are of the form $\pm \beta$, where $\beta$ is a $7$-th root of unity.
• Thanks a lot. So g is the 7th cyclotomic polynomial and its splitting field is $\Bbb{Q}(\omega)/\Bbb{Q}$ (where $\omega$ is the 7th primitive root of unity). So the Galois group must be isomorphic to the group $U_7$ of units mod 7, right? – user58289 Jul 23 '13 at 20:00
Hint: $x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$. So roots of your polynomial are roots of $x^7=-1$ except $-1$.