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  $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$

Let us replace $x$ with $z+1$ for the sake of simplicity. We have to find the sequence $\{a_n\}_{n\geq 0}$ given by:
$$ f(z)=\frac{1}{(2z-1)(z^2+z+1)}=\sum_{n\geq 0}a_n z^n \tag{1} $$
but $f(z)$ is a meromorphic function with simple poles at $z=\frac{1}{2},\,z=\omega,\,z=\omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$.
By computing the residues of $f(z)$ at such points we get a partial fraction decomposition for $f(z)$:
$$ f(z) = \frac{2}{7}\cdot \frac{1}{x-\frac{1}{2}}-\frac{1}{7}\cdot\frac{3-x-2x^2}{1-x^3} \tag{2}$$
and by performing expansions as geometric series it follows that:
$$ a_n = -\frac{4}{7} 2^n-\frac{1}{7}\cdot\left\{\begin{array}{rcl}3 & \text{if} & n\equiv 0\pmod{3}\\ -1 & \text{if} & n\equiv 1\pmod{3}\\ -2 & \text{if} & n\equiv 2\pmod{3}\end{array}\right.\tag{3}$$
that can be stated as: 
$$ a_n \text{ is the closest integer to } -\frac{2^{n+2}}{7}. \tag{4}$$
