Find the power series representation of the function I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate  $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$

Note that the geometric series is valid also in the complex field
$$
\frac{1}
{{1 - z}} = \sum\limits_{0\, \leqslant \,n} {z^{\,n} } \quad \left| \begin{gathered}
  \;z \in \;\mathbb{C}\; \hfill \\
  \,\left| z \right| < 1 \hfill \\ 
\end{gathered}  \right.
$$
and that the partial fraction decomposition can be operated also in the complex field,
so that in your case we can write
$$
\begin{gathered}
  f(z,\omega ) = \frac{1}
{{\left( {2z - 3} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} =  \hfill \\
   = \frac{1}
{{2\left( {z - 3/2} \right)\left( {z - \omega } \right)\left( {z + \omega } \right)}} =  \hfill \\
   =  - \frac{1}
{{3\left( {3/2 - \omega } \right)\left( {3/2 + \omega } \right)\left( {1 - 2/3\,z} \right)}} + \frac{1}
{{4\omega ^2 \left( {3/2 - \omega } \right)\left( {1 - z/\omega } \right)}} + \frac{1}
{{4\omega ^2 \left( {3/2 + \omega } \right)\left( {1 - \left( { - z/\omega } \right)} \right)}} \hfill \\ 
\end{gathered} 
$$
and the expression for $a_n$ follows easily.
