Find the Galois group of $x^9+x^6+x^3+1$. Question:  Let $L$ be the splitting field of $f = x^9+x^6+x^3+1 $ over $\mathbb{Q}.$ Find the Galois group $ G = \text{Gal}\left(L/\mathbb{Q}\right). $
Initially, I decomposed $f$ into its irreducible factors:
$$ f = (x^2+1)(x^4-x^2+1)(x+1)(x^2-x+1). $$
After looking up similar problems I noticed a common trick was to write 
$$ g = f\cdot(x^3-1) = x^{12}-1. $$
Let $K$ be the splitting field of $g$. Then I know that $[K : \mathbb{Q}] = \varphi(12) = 4,$ where $\varphi$ is the Euler-phi function. Moreover, we have
$$ 4=[K:\mathbb{Q}] = [K : L][L:\mathbb{Q}]. $$
I'm not sure where to go from here. This means that either $|{G}| = 2$ or $|G| = 4.$ If it's the former, then $G\cong\mathbb{Z}_2,$ but if not, I'm certainly stuck. 
Is there a more direct way to attack this problem or am I on the right track missing some links? Where should I go from here?
 A: It is easy to see that the splitting field $ L $ has degree 4 over $ \mathbf{Q} $, since it contains the subfield $ \mathbf Q(i, \sqrt{3}) $ which is of degree 4 over $ \mathbf Q $, and is a subfield of $ \mathbf Q(\zeta_{12}) $ as you have observed. Therefore, the splitting field is $ L = \mathbf Q(\zeta_{12}) $, and its Galois group over $ \mathbf Q $ is isomorphic to $ (\mathbf Z / 12 \mathbf Z)^{\times} $. It can be checked that this is actually the Klein-4 group, since there are only two groups of order 4 up to isomorphism, and this group is not cyclic.
A: the splitting field of
$X^9+X^6+X^3+1=(X+1)(X^2+1)(X^2-X+1)(X^4-X^2+1)$  is the field
obtained as composed  of the splitting field of each factors.
So the splitting field of $(X^4-X^2+1)$ is $\Bbb{Q}
(\frac{i+\sqrt{3}}{2})$, the conjugates of $\frac{i+\sqrt{3}}{2}$
are $\frac{i+\sqrt{3}}{2}$, $\frac{i-\sqrt{3}}{2}$,
$\frac{-i+\sqrt{3}}{2}$ and  $\frac{-i-\sqrt{3}}{2}$, therefore
 by the operations addition and multiplication on this numbers, in this field, we seen that  $i$  and $\sqrt{3}$ are
in  $\Bbb{Q} (\frac{i+\sqrt{3}}{2})$, it then follows that the
splitting field  $\Bbb{Q }(\frac{1+i\sqrt{3}}{2})$ of $(X^2-X+1)$,
the splitting field $\Bbb{Q} (i)$  of $(X^2+1)$ are in $\Bbb{Q}
(\frac{i+\sqrt{3}}{2})$. this extension of degree 4 over $\Bbb{Q}$
is Abelian containing at least 2 sub-extension of degree 2
therefore is not  cyclic extension, then of Galois groups
isomorphic to  $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$, as seen
on the order of its automorphisms (not trivial) that are all of
order 2. this 4 automorphisms are $\frac{i+\sqrt{3}}{2}$
associated to $\frac{\pm i\pm\sqrt{3}}{2}$
