How to do Laurent Series Expansion $f(z) =  \frac{3}{(z-1)(z-7)}$
in the domain $\{z: 1<|z|<7\}$
I remember the first step is to put $f(z)$ in partial fractions.This gives me 
$ \frac{3}{8(x-7)}$ $- \frac{3}{8(x+1)}$
What the next step ?
I'm struggling to get to the result, thanks in advance for your time

Taking the first part of the partial fraction and ignoring the coefficient,
$\dfrac{1}{z-7} = -\dfrac{1}{7}(\dfrac{1}{1-\frac{z}{7}}) = -\dfrac{1}{7}\displaystyle\sum_{n=0}^{\infty}(\frac{z}{7})^n$
Where we get the sum by noticing that we have the limit of a geometric series, and we have the formula $\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$,
 so of course, we could just put the series there (in this case, $a=1$, $r=\frac{z}{7}$). It is convergent in the annulus as $|\frac{z}{7}|<1$.
The other part is analagous, and sticking it all together will yield the Laurent series.
