Proving that a polynomial is irreducible over a splitting field

Question

As titled.Let E be the splitting field of $x^3-2$ over $\mathbb{Q}$.I know that $E = \mathbb{Q}(\sqrt[3]{2},\omega)$ where $\omega$ is the primitive root of $x^3-1$. Now,let $g(x) = x^2-7$,I want to show that g(x) is irreducible over $E$.Intuitively, I know that the roots of g(x) (namely,$\sqrt{7}$ and $-\sqrt{7}$) cannot be expressed as a combination of $\sqrt[3]{2}$ and $\omega$ in $\mathbb{Q}(\sqrt[3]{2},\omega)$.Can we prove it using Galois theory?If there's a root $\beta \in E$, then there's a subfield $\mathbb{Q}(\beta)$,with degree 2.But that doesn't seem to contradict the condition that $|\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q}| = 6$. So what's wrong with this argument? Any help will be appreciated!