Find a branch of $(z^3 - 1)^{1/3}$ which is analytic in $|z| > 1$
So we essentially want to study $\frac 13\text{Log} (1 - \frac 1{z^3} )$, the principal branch of the logarithm where $-\pi < \text{Arg} (z) < \pi $. Problems arise when $\text{Re}(z^3 - 1) \leq 0 $ and $\text{Im} (z^3 - 1) = 0$ i.e. when $z^3 - 1$ is a non-positive, real number.
Setting these equations up, how can I produce the result $|z| > 1$?
Let $$ \omega_1=1, \quad \omega_2=\frac{-1+i\sqrt{3}}{2}\quad\text{and}\quad \omega_3=\frac{-1-i\sqrt{3}}{2}. $$
A domain for $\,\sqrt[3]{1-z^3}\,$ could be $$ \Omega=\mathbb C\smallsetminus \big([0,\omega_1]\cup[0,\omega_2]\cup [0,\omega_3]\big). $$ By $[a,b]$, for $a,b\in \mathbb C$ we denote the segment connecting $a$ and $b$ in the complex plane.
Clearly $\{z:\lvert z\rvert>1\}\subset\Omega$.
In order to prove it you need the following lemma:
If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\Omega$, then $$ f(z)=\log\left(\frac{z-a}{z-b}\right), $$ is definable as holomorphic in $\Omega$.
Using this lemma, we can easily see that $\omega_1,\omega_2,\omega_3$ belong to the one and only component of $\mathbb C\smallsetminus\Omega$ and $$ \sqrt[3]{1-z^3}=(z-\omega_3)\exp\left( \frac{1}{3}\log\left(\frac{z-\omega_1}{z-\omega_3}\right)+ \frac{1}{3}\log\left(\frac{z-\omega_2}{z-\omega_3}\right) \right). $$