How to prove that the polynomial $x^2 + x + 1$ is irreducible over $\mathbb Q(\sqrt[3]{2})$?

I am trying to prove that the polynomial $x^2 + x + 1$ is irreducible over $\mathbb Q(\sqrt[3]{2}).$

This is my guess:

$x^2 + x + 1$ is reducible over $\mathbb Q(\sqrt[3]{2})$ iff $\xi \in \mathbb Q(\sqrt[3]{2})$

where $\xi = \frac{-1 + i \sqrt{3}}{2}$ or $\xi = \frac{-1 - i \sqrt{3}}{2}$ but this is not true as $\xi$ is a complex number.

Is my justification correct? if not, please correct me.

Solution 1:

$[\mathbb Q(\sqrt[3]{2}):\mathbb Q]=3$, hence there is no intermediate extension of degree $2$, and it's sufficient to show $x^2+x+1$ is irreducible over $\mathbb Q$.

Another way: the two roots of $x^2+x+1$ are all imaginary, hence neither of them is included in $\mathbb Q(\sqrt[3]{2})\subset\mathbb R$.