This is an exam question that i'm trying to figure out.
Apart from the title, there is a note added to the question that says that: The branch of $\sqrt[3]{}$ is the one that has $\sqrt[3]{7} \in \mathbb{R}$
For the integral I tried using the Residue at infinity using the usual substitution. I get that $$g(w) = \frac{1}{w^3}\sqrt[3]{1-w^3}$$ is analytic in $|z|=\frac{1}{2}$ except at zero where $w=0$ is an order $3$ pole. So the residue at zero is: $$res_0(g) = \frac{(\sqrt[3]{1-w^3})''(0)}{2!} = 0$$ Therefore the integral equals $0$.
Although all the above, I feel something is wrong cause I did not use the information about the branch at all, and frankly I have no idea how to use it. I know that there are three branch points, namely: $1, e^{\frac{2}{3}i\pi},e^{-\frac{2}{3}i\pi}$ so I guess a branch cut will have to be two rays coming out of $1$ and passing through the other two?
But still I fail to see how the fact that the branch has to be one so that $\sqrt[3]{7} \in \mathbb{R}$ helps me here, is it possible to have a branch where it isn't so? Perhaps I still don't really understand branches as well as I thought I did.
Any thoughts or input on this will be greatly appreciated. Thank you!