I found this resource by Dr. Keith Conrad showing how to compute the basis of $\mathbb{Q}(\sqrt[3]{2},\omega)$ over $\mathbb{Q}$ (where $\omega$ denotes the third primitive root of unity): Keith Conrad
In class, when we were taught to compute the basis of an extension, we were told to find elements in the extension that were not in the fixed field. For example, the basis of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $\{a_1+a_2\sqrt{2}\}$ (since $\sqrt{2}^2=2\in\mathbb{Q}$), where each $a_j\in\mathbb{Q}$. Similarly, the basis of $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}$ is $\{a_1+a_2\sqrt[4]{2}+a_3\sqrt{2}+a_4\sqrt[4]{8}\}$ (since $\sqrt[4]{2}^2,\sqrt[4]{2}^3 \not\in\mathbb{Q}$, and $\sqrt[4]{2}^4 = 2 \in \mathbb{Q}$). However, I am having a bit of trouble understanding the basis of $\mathbb{Q}(\omega)$. I know that $\lbrack \mathbb{Q}(\omega): \mathbb{Q} \rbrack=2$, so the basis should have just two elements; however, would the basis not be $\{a_1+a_2\omega+a_3\omega^2\}$? And this basis has three elements.
Moreover, for the extension $\mathbb{Q}(\sqrt[3]{2},\omega)$, I computed the basis $\{a_1+a_2\sqrt[3]{2}+a_3\omega\sqrt[3]{2}+a_4\omega^2\sqrt[3]{2}+a_5\sqrt[3]{4}+a_6\omega\sqrt[3]{4}+a_7\omega^2\sqrt[3]{4}+a_8\omega+a_{9}\omega^2\}$. I know that $\lbrack \mathbb{Q}(\sqrt[3]{2},\omega): \mathbb{Q} \rbrack = 6$, so I expect a basis with 6 elements, which I found how to compute in Dr. Conrad's notes. So how would I know that the basis that I computed is not minimal? Because the elements in the basis that I computed do not seem to be linearly dependent with one another, and do span $\mathbb{Q}(\sqrt[3]{2},\omega)$.
Regarding your question: so the basis should have just two elements; however, would the basis not be $\{a_1+a_2\omega+a_3\omega^2\}$?
No. You have $\omega^2 +\omega+1=0$. So the elements $1,\omega, \omega^2$ are not linearly independent. $(1,\omega)$ is a basis of $\mathbb Q(\omega)/\mathbb Q$.
Using the telescopic basis theorem, you get that a basis of $\mathbb Q(\sqrt[3]{2}, \omega)/\mathbb Q$ is
$$(1, \sqrt[3]{2}, \sqrt[3]{4}, \omega, \sqrt[3]{2}\omega, \sqrt[3]{4}\omega).$$