Galois group of $x^6-9$ $f = x^6-9 = (x^3-3)(x^3+3)$
Let $L_f$ be splitting field therefore $L_f = \mathbb{Q}[\sqrt[3]{3},e^{\frac{2\pi i}{3}}]$, $[L_f:\mathbb{Q}] = 9$.
Also $Gal\space x^3±3/\mathbb{Q} = S_3$ and $Gal\space f/\mathbb{Q}$-orbits is roots of $x^3-3$ and $x^3+3$ therefore, i think, that $G = Gal\space f/\mathbb{Q} = S^3\times S^3$. But $|G| = 36 \ne [L_f:\mathbb{Q}]$. What's wrong?
 A: A few issues:
Firstly $e^{2\pi i/3}$ is a root of the polynomial $X^2+X+1$, so the degree of $L_f$ over $\mathbb Q$ will be $6$.
Secondly, as quid has mentioned, both $X^3\pm3$ have the same splitting fields. Hence, the splitting field of $f$ is the same as the splitting field of either of those polynomials, so you can just find the Galois group of one of them. 
With these two facts you should be able to find that the group is $S_3$. 
A: A problem is that the two fields generated by the roots of $x^3-3$ and $x^3 +3$ are not at all independent. 
Indeed, the splitting fields of $x^3-3$ and $x^3 +3$ are the same. 
A: For the sake of completeness, I am posting here an entire solution (taken from http://www.polishedproofs.com/galois-group-of-x6-9/)
Notice that $h:=x^6-9=(x^3-3)(x^3+3)$. Define
$$f(x):=x^3-3$$
$$g(x):=x^3+3$$
Thus, $h=fg$. Notice that both $f$ and $g$ are irreducible by Eisenstein criterion with $p=3$, since $p| \pm 3, p \nmid 1$, and $p^2 \nmid \pm 3 $. Also, both $f$ and $g$ are separable, since they are polynomials over a field of characteristic $0$. Notice that the roots of $f$ are
$$3^{\frac{1}{3}}, 3^{\frac{1}{3}} \omega, 3^{\frac{1}{3}} \omega^2$$
where $\omega$ is the $3$rd primitive root of unity. Similarly, roots of $g$ are
$$-3^{\frac{1}{3}}, -3^{\frac{1}{3}} \omega, -3^{\frac{1}{3}} \omega^2$$
Therefore, by adjoining the roots of $f$ we automatically adjoin roots of $g$. Thus, it is enough to find the Galois group of $f$.
Since $f$ is an irreducible, monic, separable polynomial of degree $3$ and $\mathbb{Q}$ has characteristic different from $2$ and $3$, we can find its discriminant, namely
$$D=\triangle^2 = -4(0)^2-27(-3)^2=-243$$
which is not a square in $\mathbb{Q}$. Therefore, the Galois group of $f$ (and thus $h$) is S$_3$.
