Evaluate the complex integral of function 
Use the residue theorem to evaluate $\int_\gamma \frac{z^5}{1-z^3}dz$
  where $\gamma$ is the circle $|z|=2$.

I have that $z_0=1$ is a singularity point and taking $g(z)=z^5$ and $h(z)=1-z^3$ and knowing that $g(z_0)\neq 0$ , $h(z_0)=0$ and $h'(z_0)\neq 0$ then $f(z)=\frac{g(z)}{h(z)}$ have simple pole at $z_0$ and  $$Res(f;z_0)=\frac{g(z_0)}{h'(z_0)}=-\frac{1}{3}$$
then $\int_\gamma \frac{z^5}{1-z^3}dz=2\pi i* Res(f;z_0)=-\frac{2\pi i}{3}$
but the answer is $-2\pi i$

Here is an approach that avoids factorization. Start by noticing
$$
\frac{z^5}{1-z^3}=\frac{z^2}{1-z^3}-z^2
$$
Since there are no singularities outside of the unit circle, for any $R\ge2$, we have
$$
\begin{align}
\int_{|z|=2}\frac{z^5}{1-z^3}\,\mathrm{d}z
&=\int_{|z|=2}\frac{z^2}{1-z^3}\,\mathrm{d}z\\
&=\int_{|z|=R}\frac{z^2}{1-z^3}\,\mathrm{d}z
\end{align}
$$
Furthermore, noticing that $\frac{z^2}{1-z^3}=-\frac1z\frac1{1-1/z^3}$ is very close to $-\frac1z$ when $|z|$ is large, we add the integral of $\frac1z$ in hopes that it will cancel.
$$
\begin{align}
\left|\,\int_{|z|=R}\frac{z^2}{1-z^3}\,\mathrm{d}z+\int_{|z|=R}\frac1z\,\mathrm{d}z\,\right|
&\le\sup_{|z|=R}\left|\,\frac{z^2}{1-z^3}+\frac1z\,\right|2\pi R\\
&\le\sup_{|z|=R}\left|\,\frac1{(1-z^3)z}\,\right|2\pi R\\
&\le\frac{2\pi}{R^3-1}
\end{align}
$$
Since $R$ is arbitrary, the sum of the integrals on the left must be $0$. Therefore,
$$
\begin{align}
\int_{|z|=2}\frac{z^5}{1-z^3}\,\mathrm{d}z
&=\int_{|z|=2}\frac{z^2}{1-z^3}\,\mathrm{d}z\\
&=-\int_{|z|=2}\frac1z\,\mathrm{d}z\\[6pt]
&=-2\pi i
\end{align}
$$
