Splitting field of $x^3-2$ over $\Bbb Q$ are distinct but isomorphic What I'm trying to solve is:
Prove that given the polynomial $x^3-2$ over $\Bbb Q$, its splitting fields $\Bbb Q(x_1)$ , $\Bbb Q(x_2)$ , $\Bbb Q(x_3)$ are isomorphic but distinct (where $x_1$, $x_2$, $x_3$ are the roots of the polynomial).
What I've tried so far:
I managed to prove they are isomorphic, but how do I go about the distinction between them? Thanks to everyone in advance.
 A: Let the zeros of $f(x)=x^3-2$ be $x_1=\root3\of2$, $x_2=\rho x_1$, and $x_3=\rho^2 x_1$, where $\rho=e^{2\pi i/3}$ is a primitive cube root of unity. Note that $x_1+x_2+x_3=0$, so any field containing two of the three zeros also contains the third.
$x_1$ is real, and of degree three over the rationals.
$x_2$ is not real, so it isn't in $K={\bf Q}(x_1)$, so $K$ is not a splitting field for $f$. $x_2$ is a zero of $(x^3-2)/(x-x_1)=x^2+x_1x+x_1^2$ so it's of degree two over $K$ (not degree one, since we know it's not in $K$).
$L=K(x_2)$ contains all the zeros of $f$, and no smaller field does, so $L$ is a splitting field for $f$. Since $L$ has degree two over $K$, and $K$ has degree three over the rationals, $L$ has degree six over the rationals.
${\bf Q}(x_2,x_3)$ contains all the zeros of $f$, so it contains (in fact, is) a splitting field for $f$, so it has degree at least (in fact, exactly) six over the rationals, so $x_3$ isn't in ${\bf Q}(x_2)$ (if it were, then ${\bf Q}(x_2,x_3)$, of degree at least six, would equal ${\bf Q}(x_2)$, of degree three, which is nonsense). Therefore, ${\bf Q}(x_2)$ and ${\bf Q}(x_3)$ are distinct.
A: Let $x_1$ be the real root. That's obviously distinct from the other two. Suppose $\mathbb{Q}(x_2)=\mathbb{Q}(x_3)$. Then, $x_2,x_3$ are in the same extension, w.l.o.g. let it be $\mathbb{Q}(x_2)$. Since $x_1x_2x_3=2$, that implies $x_1$ is also in $\mathbb{Q}(x_2)$. So $\mathbb{Q}(x_1)\subsetneq \mathbb{Q}(x_2)$. Which is impossible.
