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?
 A: Hint: From $z^3-1=(z-1)(z^2+z+1)$ look at the following function $g(z)=\frac{1}{z^2+z+1}$. Then from Cauchy's integral formula, since $1 \in \{ z : | z − 1| < 1\}$
$$g(1)=\frac{1}{2\pi i}\int_{|z−1|=1} \frac{g(z)}{z-1}dz=\frac{1}{2\pi i}\int_{|z−1|=1} \frac{1}{z^3-1}dz$$
A: The only pole of $f(z)$ on the circle is $z=1$
According to residue theorem
$$\oint _{\gamma }f(z)\,dz=2\pi i \operatorname {Res} (f,1)$$
And we have $\operatorname {Res} (f,1)=\lim_{z\to 1} \, \dfrac{z-1}{z^3-1}=\lim_{z\to 1} \, \dfrac{1}{z^2+z+1}=\dfrac{1}{3}$
Thus $\oint _{\gamma }f(z)\,dz=\dfrac{2 i \pi }{3}$
Hope this helps
A: 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*}

A: Determine which roots of the denominator are contained within the given curve $|z - 1| = 1$, find the residues at those roots, and calculate using the Residue Theorem.
