Since $\sqrt[3]{2}, i \in \mathbb{Q}(\sqrt[3]{2}, i)$, their product $\sqrt[3]{2}i \in \mathbb{Q}(\sqrt[3]{2}, i)$, which implies $\mathbb{Q}(\sqrt[3]{2}i) \subset \mathbb{Q}(\sqrt[3]{2}, i)$. However, I am told that this is a tricky question. I have stared at this question for a while but still cannot figure out where the tricky part is? Can someone point it out to me? Thanks in advance!
EDIT: It seems that there is a typo in the book and the tricky part is showing the inclusion is the other way around, i.e. $\mathbb{Q}(\sqrt[3]{2}, i) \subset \mathbb{Q}(\sqrt[3]{2}i)$.
My thoughts: Since $(\sqrt[3]{2}i)^3 = -2i$, $i \in \mathbb{Q}(\sqrt[3]{2}i)$. So $\sqrt[3]{2} = (1/i)\cdot \sqrt[3]{2}i \in \mathbb{Q}(\sqrt[3]{2}i)$. So $\mathbb{Q}(\sqrt[3]{2}, i) \subset \mathbb{Q}(\sqrt[3]{2}i)$. Since it is clear that $\mathbb{Q}(\sqrt[3]{2}i) \subset \mathbb{Q}(\sqrt[3]{2}, i)$. The two fields are actually equal.
It seems not so clear why this is a "tricky" question. Perhaps the following question was intended (but of course I cannot say):
Show that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$.
This question is quite a popular question, so it has been solved here a few times, see also here, or here:
How can I prove that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$?
Edit: Another possibility is to show that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}i)$, and furthermore, that $\mathbb{Q}(\sqrt[3]{2},\omega)\supsetneq\mathbb{Q}(\sqrt[3]{2}\omega)$, arising by the splitting field of $X^3-2$.