First way: Using Cauchy's differentiation formula $f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\,dz$ we get that$\oint_{|z-1|=2}\frac{1}{(z-2)^3}\,dz=\left.\frac{2\pi i}{2!}\,\frac{\mathcal d^2}{\mathcal d z^2}(1)\,\,\right|_{z=2}=0$
Second way: Making first the translation $\,z\to z+1\,$ , we take the circle of radius $\,2\,$ centered at the origin and $\,(z-1)^3\,$ in the denominator, getting $I:=\oint_{|z|=2}\frac{1}{(z-1)^3}\,dz$
Now, we put $\,z=2e^{i\theta}\,\,,\,0\leq\theta\leq 2\pi\Longrightarrow dz=2ie^{i\theta}\,$ , so we get$I=\int_0^{2\pi}\frac{2ie^{i\theta}d\theta}{\left(2e^{i\theta}-1\right)^3}=\int_0^{2\pi}\frac{(2e^{i\theta})}{\left(2e^{i\theta}-1\right)^3}=\left.-\frac{1}{2}\frac{1}{(2e^{i\theta}-1)^2}\right|_0^{2\pi}=$$=-\frac{1}{2}\left(\frac{1}{(2-1)^2}-\frac{1}{(2-1)^2}\right)=0$