$\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ Galois group How do I calculate the degree of $\mathbb{Q}(\sqrt[3]{2}, \zeta_{9})$ over $\mathbb{Q}$. Should it be 18, as $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$, and $[\mathbb{Q}(\zeta_{9}):\mathbb{Q}] = 6$? 
However $(\sqrt[3]{2})^{3} \in \mathbb{Q}(\zeta_{9})$, how this affect the calculation?
Thanks

How about doing it the other way around:
Certainly $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ is $3$, so we aim to compute $n=[L:K]=|Gal(L/K)|$, where $L = \mathbb{Q}(\sqrt[3]{2},\zeta_9)$ and $K=\mathbb{Q}(\sqrt[3]{2})$. The extension is cyclotomic, so certainly is Galois. 
The minimal polynomial of $\zeta_9/K$ divides $\varphi_9(X)$, so we know that $n \leq6$. Now note that $L$ contains the subextension $\mathbb{Q}(\zeta_9)/\mathbb{Q}$ of degree $6$, so $3$ divides $n$. But since $K$ is real and $L/K$ is Galois, complex conjugation gives an order $2$ element inside $Gal(L/K)$, so $2$ divides $n$. Thus $n = 6$.
