Answer
$\frac{-10x^2+27x-14}{(x-1)^3(x+2)}=\frac{4}{x+2}+\frac{-4}{x-1}+\frac{2}{(x-1)^2}+\frac{1}{(x-1)^3}$,
Work Step by Step
$\frac{-10x^2+27x-14}{(x-1)^3(x+2)}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{(x-1)^2}+\frac{D}{(x-1)^3}$,
$A(x-1)^3+B(x-1)^2(x+2)+C(x-1)(x+2)+D(x+2)$,
$A(x^3-3x^2+3x-1)+B(x^3-3x+2)+C(x^2+x-2)+D(x+2)$,
$(A+B)x^3+(-3A+C)x^2+(3A-3B+C+D)x+(-A+2B-2C+2D)$,
$\begin{cases}
& A & +B=0\\
& -3A & +C = -10\\
3A & -3B & +C +D=27\\
-A & +2B & -2C +2D= -14
\end{cases}$
Multiplying Equation $4$ by $3$ and adding it to Equation $3$.
$\begin{cases}
-3A +6B -6C +6D=-42\\
3A -3B +C +D=27\\
-- -- -- -- -- -- -- --\\
0A +3B -5C +7D=-15
\end{cases}$
Adding Equation $2$ and Equation $3$.
$\begin{cases}
& -3A +C=-10\\
& +3A -3B +C +D=27\\
& -- -- -- -- -- -- -- --\\
& 0A -3B +2C +D=17
\end{cases}$
Multiplying Equation $1$ by $3$ and adding it to Equation $2$.
$\begin{cases}
& 3A +3B=0\\
& -3A +C=-10\\
& -- -- -- -- --\\
& 0A +3B +C=-10
\end{cases}$
Adding the Addition result of Equation $4$, Equation $3$ and the Addition result of Equation $2$, Equation $3$
$\begin{cases}
& 3B -5C +7D=-15\\
& -3B +2C +D=17\\
& -- -- -- -- -- --\\
& -3C +8D=2
\end{cases}$
Adding the Addition result of Equation $1$, Equation $2$ and Equation $2$, Equation $3$.
$\begin{cases}
& -3B +2C +D=17\\
& 3B +C=-10\\
& -- -- -- -- -- &\\
& 3C +D=7
\end{cases}$
Adding the Addition result...
$\begin{cases}
& -3C+8D=2\\
& 3C+D=7\\
& -- -- -- --\\
& +9D=9
\end{cases}$
thus, $D=1$, Substituting back into the Equation, $3C+1=7, C=2$.
Substituting back into the Equation, $3B+2=-10, B=-4$.
Substituting back into the Equation, $A-4=0, A=4$.
Therefore,
$\frac{-10x^2+27x-14}{(x-1)^3(x+2)}=\frac{4}{x+2}+\frac{-4}{x-1}+\frac{2}{(x-1)^2}+\frac{1}{(x-1)^3}$,
