Answer
$\frac{9x^2-9x+6}{(x-2)(x+2)(2x-1)}=\frac{2}{x-2}+\frac{3}{x+2}+\frac{-1}{2x-1}$,
Work Step by Step
$\frac{9x^2-9x+6}{2x^3-x^2-8x+4}$, factorising $2x^3-x^2-8x+4=x^2(2x-1)-4(2x-1)=(x^2-4)(2x-1)=(x-2)(x+2)(2x-1)$:
$\frac{9x^2-9x+6}{(x-2)(x+2)(2x-1)}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{2x-1}$,
$9x^2-9x+6=A(x+2)(2x-1)+B(x-2)(2x-1)+C(x^2-4)$,
$9x^2-9x+6=A(2x^2+3x-2)+B(2x^2-5x+2)+C(x^2-4)$,
$9x^2-9x+6=(2A+2B+C)x^2+(3A-5B)x+(-2A+2B-4C)=9x^2-9x+6$,
$\begin{cases}
& 2A & +2B & +C= 9&\\
& -2A & +2B & -4C= 6&\\
& 3A & -5B & +0C =-9&\\
\end{cases}$
Adding the Equation $1$ and Equation $2$,
$\begin{cases}
& 2A & +2B & +C =9\\
& -2A & +2B & -4C = 6\\
&-- -- & -- -- & -- -- &\\
& +0A & 4B & -3C = 15
\end{cases}$
Multiplying Equation $2$ by $\frac{3}{2}$ and Adding it to Equation $3$,
$\begin{cases}
& -3A & +3B & -6C = 9\\
& 3A & -5B & +0C = -9\\
& -- -- & -- -- & -- -- &\\
& 0A & -2B & -6C = 0
\end{cases}$
Multiplying the result of the second Addition by $2$ and Adding it to the result of the first addition:
$\begin{cases}
& 4B & -3C = 15\\
& -4B & -12C =0\\
& -- -- & -- --& \\
& 0 & -15C = 15
\end{cases}$
thus, $C=-1$
Substituting back $C=-1$ into the equation, $4B+3=15, B=3$. Substituting $B$ into the third Equation, $3A-15=-9, A=2$.
Therefore,
$\frac{9x^2-9x+6}{(x-2)(x+2)(2x-1)}=\frac{2}{x-2}+\frac{3}{x+2}+\frac{-1}{2x-1}$,
