Deformation Theorem:
Let $\gamma$ be a simple closed curve. Let $D(a, r)$ be an open disc lies inside $\gamma$.
Let $f$ be an analytic function on and inside $\gamma$ except maybe at $a$.
Then $\int_{\gamma} f(z) dz = \int_{|z-a|=r} f(z) dz$
In our question, $f(z) = \frac{1}{z^3 - 1}$, which is analytic everywhere except at the three points labeled on the graph, namely, 1, $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Let A, B, C be three complex numbers such that $f(z) = \frac{A}{z-1} + \frac{B}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{C}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$
Solve for A,B,C we get
$$A = \frac{1}{3}$$ $$B = \frac{2}{-3\sqrt{3}i-3}$$ $$C = \frac{2}{3\sqrt{3}i-3}$$
So, $f(z) = \frac{\frac{1}{3}}{z-1} + \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} + \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)}$
Let $\gamma$ be the blue curve.
The Deformation theorem implies that $$\int_{\gamma} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{1}{z^3 - 1} dz$$ for some $r>0$.
$$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz
=\int_{|z-1|=r} \frac{\frac{1}{3}}{z-1}
+ \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)}
+ \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz
= \int_{|z-1|=r}\frac{\frac{1}{3}}{z-1} dz
+\int_{|z-1|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz
+ \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz $$
Since $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$ are outside of the unit circle $|z-1|=r$, then $\int_{|z-1|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz = 0$ and $\int_{|z-1|=r} \int_{|z-1|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz =0$ by Cauchy's theorem.
Therefore, $\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-1|=r} \frac{\frac{1}{3}}{z-1} dz = \frac{1}{3}\int_{|z-1|=r} \frac{1}{z-1} dz = \frac{1}{3}\cdot 2\pi i = \frac{2\pi i}{3}$.
Thus, the integral of $f$ on the blue curve is $ \frac{2\pi i}{3}$.
Similarly, for the pink curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz =\int_{|z-(-\frac{1}{2} + i\frac{\sqrt{3}}{2})|=r} \frac{\frac{2}{-3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)} dz = \frac{2}{-3\sqrt{3}i-3} \cdot 2\pi i = \frac{4\pi i}{-3\sqrt{3}i-3}$
For the brown curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{1}{z^3 - 1} dz = \int_{|z-(-\frac{1}{2} - i\frac{\sqrt{3}}{2})|=r} \frac{\frac{2}{3\sqrt{3}i-3}}{z-\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)} dz = \frac{2}{3\sqrt{3}i-3} \cdot 2 \pi i = \frac{4 \pi i}{3\sqrt{3}i-3}$
For the green curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4\pi i}{-3\sqrt{3}i-3} $
For the orange curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$
For the red curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4 \pi i}{3\sqrt{3}i-3}$
For the purple curve,
$\int_{|z-1|=r} \frac{1}{z^3 - 1} dz = \frac{2\pi i}{3} + \frac{4\pi i}{-3\sqrt{3}i-3} + \frac{4 \pi i}{3\sqrt{3}i-3}$
This concludes the question.
