Compute this integral: 
$$\int\_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$

my solution is I used derivative of Cauchy integral formula, which is 
$$f^{(n)}(z\_0) = \frac{n!}{2\pi i}\int \frac{f(z\_0)}{(z-z\_0)^{n+1}}$$
Then, I got
$$\int\_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$ 
$$ = \int\_{|z|=1} \frac{(z-2)^3}{(2z-1)^3}dz$$
$$ = \frac{2 \pi i}{2!}\left(\frac{1}{2} - 2\right)^3 = \frac{27\pi i}{8}$$

That's what I get from the Cauchy integral formula. So, am I correct for this approach. If not, can someone please show me how to get the right one? Thank you