Let $K$ be an field extension of $F$. Then I know there is a theorem that says: $K$ is a splitting field of some polynomial over $F$ iff $K$ over $F$ is finite and normal. (Here the normal means every polynomial in $F[x]$ has a root then have all roots.)
Now, $\Bbb Q(\sqrt[3]{2})$ does not have all roots of $x^3-2$, so it is not normal. Is it possible that $\Bbb Q(\sqrt[3]{2})$ has all roots of some polynomial $p(x)$, but these roots are all not in $\Bbb Q$?
On
No.
First, such a polynomial $p(x)$ would have to be irreducible over $\mathbb Q$. If it wasn't, it would split into factors at least one of which would be linear. This would yield a root in $\mathbb Q$.
If $p(x)$ is such an irreducible polynomial, then we have that $\mathbb Q (^3 \! \sqrt2)$ contains the splitting field of $p(x)$. But this extension has no intermediate extensions. (By multiplicity of degrees of extensions).
So $\mathbb Q (^3 \! \sqrt2)$ becomes the splitting field for an irreducible polynomial over $\mathbb Q$, which makes it a normal extension. But we know that this isn't the case - The polynomial $x^3 - 2$ has only one root here, out of the three possible.
They'd all have to be $\notin \mathbb Q$, as otherwise $p$ would be reducible.