How to show $\sqrt[3]{X-i}\notin \mathbb{C}(X,\sqrt[3]{X+i})$ I'm trying to show $\sqrt[3]{X-i}\notin \mathbb{C}(X,\sqrt[3]{X+i})$. But this is harder than I expected.
Is there any easy way to show this?
 A: Its easier for me to give my solution over a general field.
Let $k$ be a field with a cube root of unity $\zeta$. In our case $k=\mathbb{C}(x)$
Let $a,b \in k$ and assume that $\sqrt[3]{b} \in k(\sqrt[3]{a})$. Now the congugates of 
$\sqrt[3]{a}$ are $\sqrt[3]{a},\zeta\sqrt[3]{a},\zeta^2\sqrt[3]{a}$ so in fact 
$\text{Tr} (\sqrt[3]{a})=0$.
If we have 
$$\sqrt[3]{b}=\alpha+\beta \sqrt[3]{a}+\gamma(\sqrt[3]{a})^2$$ then taking the trace of both sides, gives $3\alpha=0$ so we really have 
$$\sqrt[3]{b}=\beta \sqrt[3]{a}+\gamma(\sqrt[3]{a})^2$$ let $\sigma $ be the automorphism such that 
$\sigma(\sqrt[3]{a})=\zeta\sqrt[3]{a}$ so we have either 
$$\sigma(\sqrt[3]{b})=\sqrt[3]{b} \ \ \text{or} \ \ \zeta\sqrt[3]{b} \ \  \text{or} \ \ \zeta^2\sqrt[3]{b}$$
Say $\sigma(\sqrt[3]{b})=\zeta\sqrt[3]{b}$ then by applying $\sigma$ to the above equation we have 
$$\zeta\sqrt[3]{b}=\zeta\beta \sqrt[3]{a}+\zeta^2\gamma(\sqrt[3]{a})^2$$, however if the multiply the same equation by $\zeta$ and subtract we get $\gamma=0$. (the first case is seen to be impossible and the third gives a similar result)
So we have now
$$\sqrt[3]{b}=\beta \sqrt[3]{a}$$ and cubing we have 
$$\frac{b}{a}=\beta^3 $$
So to prove our result we need only remark that 
$\frac{x-i}{x+i}$ is not the third power of a rational function in $\mathbb{C}(x)$
