Complex Integration using Cauchy's Theorem The problem is the integration of 
$$I=\int_{\left\lvert z-1\right\rvert=1} f(z) dz$$
where 
$$f(z)=\frac{1}{z^3-1}$$
and the path goes $1$ loop in positive direction.
I tried to solve the problem using Cauchy's Theorem by finding $z$ that makes $f(z)$ denominator be $0$. That was $z=1$.
And I got struck. 
I think the integral is needed to be treat somehow so that 
$$\int_{C_a}\frac{1}{(z-\alpha)^n}dz = 2{\pi}i \text{ when } n=1$$ 
can be used. 
My question is how should I continue with the integral?

Note:  The hint of @rtybase is essentially the answer to OPs question. Let's recall Cauchy's Integralformula:
If $a$ is in the interior  of $\gamma=\{z:|z-1|=1\}$ and a function $g$ is holomorphic in a region which contains the closure of the interior of $\gamma$, then
\begin{align*}
g(a)=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-a}\,dz
\end{align*}

We have the following situation:
The function $f(z)=\frac{1}{z^3-1}$ has three simple poles at $z_0=1$ and  $z_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$. The pole $z_0=1$ is in   the interior of $\gamma$. We can write $f(z)$ as
  \begin{align*}
f(z)=\frac{1}{z^3-1}=\frac{1}{(z-1)(z^2+z+1)}
\end{align*}
  and observe: Since $g(z)=\frac{1}{z^2+z+1}$ is holomorph in the interior  of  $\gamma$ we obtain
  \begin{align*}
g(1)&=\frac{1}{2\pi i}\oint_{\gamma} \frac{g(z)}{z-1}\,dz\\
&=\frac{1}{2\pi i}\oint_{\gamma} \frac{1}{(z-1)(z^2+z+1)}\,dz\\
&=\frac{1}{2\pi i}\oint_{\gamma} f(z)\,dz\\
\end{align*}
We finally conclude
  \begin{align*}
\oint_{\gamma} f(z)\,dz=2\pi i\, g(1)=\frac{2\pi i}{3}
\end{align*}

