I know that the field extension $\mathbb{Q(\sqrt[3]{2},\zeta)}$, where $\zeta = -\frac{1}{2}+\frac{1}{2}\sqrt{-3}$ can be represented as a six dimensional vector space over $\mathbb{Q}$ with basis $\{1, \sqrt[3]{2}, \sqrt[3]{4}, \zeta, \sqrt[3]{2}\cdot \zeta, \sqrt[3]{4} \cdot \zeta \}$. I am experimenting with a matrix representation of this vector space.
Is $\{1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt{-3}, \sqrt[3]{2}\cdot \sqrt{-3}, \sqrt[3]{4} \cdot \sqrt{-3} \}$ also a basis of the vector space representing $\mathbb{Q(\sqrt[3]{2},\zeta)}$?
If not, then why not?