Splitting Field of $x^6-6x^3+7$ Given the polynomial $p(x)=x^6-6x^3+7 \in \mathbb Q$, find its splitting field $\mathbb F\subset \mathbb C$ and the Galois group of the extension $\mathbb F /\mathbb Q$.
In fact, the exercise asked to show that $p$ is irreducible, then the degree and a basis of the splitting field as a vector space over $\mathbb Q$, and finally the Galois group of the extension.
The six complex roots are $\zeta_3^k \cdot \sqrt[3]{3\pm \sqrt2}$ with $\zeta_3$ a primitive third root of 1 and $k=0,1,2$. 
I've worked out a clumsy solution, so I was thinking to myself:


*

*Is there a tricky/easy/obvious(!) way to show that $p$ is irreducible over $\mathbb Q$?

*Is there an easy way to show that $[\mathbb Q(\sqrt 2, \sqrt[3]{3\pm\sqrt2}):\mathbb Q(\sqrt 2, \sqrt[3]{3+\sqrt2})]=3$? 
It seems to me that there's no way to avoid this step or a similar one, like $[\mathbb Q(\sqrt 2, \sqrt[3]{3-\sqrt2}),\mathbb Q(\sqrt 2, \sqrt[3]7)]=3$ or else (you could add $\zeta_3$ just after $\sqrt 2$..). 
I'm quite sure that $[\mathbb F:\mathbb Q]=36$, but the only non trivial fact in the proof is the one I've just mentioned. Anyway, I could be missing something.. 

*Claim: the Galois group is $S_3\times S_3$, that is $D_3\times D_3$ (dihedral). Am I right? How can I prove it?

 A: There is no "big hammer" for proving irreducibility.  Eisenstein's criterion sometimes works, but does not do so for the problem polynomial nor for the additional polynomial in your comment.
Tricky/obvious solution:  $x$ only appears as powers of its cube, so consider the substitution $x^3 \rightarrow y$, yielding $y^2-6y+7$.  This has discriminant 36-28 = 8 which is not a square, so $y^2-6y+7$ does not split over $\mathbb{Q}$.  Let $r \in \mathbb{C}$ be (either) root of this equation.  Then $y^2-6y+7$ does split over $\mathbb{Q}(r)$.
Index = 3:  This is easy since you know how to split the cube, yielding three roots in $x$ per root in $y$.
Solved this way, you get a sequence of two extensions from which you should be able to write down your Galois group.
A: Partial answer - only considering irreducibility at this time.
Let us first reduce the coefficients modulo $5$, and show that the polynomial is irreducible in the ring $\Bbb{F}_5[x]$. I first show that the splitting field over $\Bbb{F}_5$ is of degree six. Obviously the element $3+\sqrt{2}\in\Bbb{F}_{25}$ needs to be in the splitting field as do its cube roots. Note that $2$ is not a quadratic residue
modulo five, so $\Bbb{F}_5[\sqrt2]=\Bbb{F}_{25}$.
Claim. $3+\sqrt2$ is not a cube in $\Bbb{F}_{25}$.
Proof. Every non-zero element $\beta$ of $\Bbb{F}_{25}$ satisfies the equation $\beta^{24}=1$. So assuming contrariwise that $\beta^3=3+\sqrt2$, we would get
$$
1\equiv(\beta^3)^8=((3+\sqrt2)^2)^4\equiv(1+\sqrt2)^4=(3+2\sqrt2)^2\equiv2+2\sqrt2\pmod 5,
$$
which is a contradiction. QED
This shows that the order of $3+\sqrt2$ in $\Bbb{F}_{25}^*$ is a multiple of three.
Therefore the order of its cube root (in some extension field of $\Bbb{F}_5$ is a multiple of nine. The smallest extension field of $\Bbb{F}_5$ that has elements of order nine is
$\Bbb{E}=\Bbb{F}_{5^6}$. The irreducibility of $p(x)$ over $\Bbb{F}_5$ follows from this.
If $p(x)$ were reducible over $\Bbb{Q}$ it would factor in $\Bbb{Z}[x]$ and, by reducing that factorization modulo five, also in $\Bbb{F}_5[x]$. Hence we can conclude that $p(x)$ is irreducible over the rationals.
