How can I prove that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$? It is easy to show that $\mathbb{Q}(\sqrt[3]{2}+i)\subseteq \mathbb{Q}(\sqrt[3]{2},i)$. 
But how I can show that $\mathbb{Q}(\sqrt[3]{2},i)\subseteq\mathbb{Q}(\sqrt[3]{2}+i)$?
I can't find a way to express $\sqrt[3]{2}$ in terms of $\sqrt[3]{2}+i$.

Consider numbers of the form
$$x_0+x_1a+x_2a^2+x_3i+x_4ai+x_5a^2i$$
where $\sqrt[3]2=a$ and $x_i\in\mathbb Q$. It is easily shown that these numbers form a vector space $V$ under addition and are closed under multiplication.
Now consider the powers of $\sqrt[3]2+i=a+i=z$ from $z^0=1$ to $z^5$. Certainly all these numbers are of the form above, and the six powers of $z$ are linearly independent in $V$, so they are a basis. By inverting this basis, it is possible to derive an expression for $\sqrt[3]2$ in terms of powers of $z$, showing that it is in $\mathbb Q(\sqrt[3]2,i)$, and similarly for $i$.
Specifically,
$$\sqrt[3]2=\frac1{22}(91+100z-78z^2+40z^3-9z^4+12z^5)$$
