Answer
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
Work Step by Step
We are given the fraction:
$\dfrac{x^2-6x+3}{(x-2)^3}$
We will decompose the fraction: as the denominator has the form $(x-a)^k$, we have:
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}+\dfrac{C}{(x-2)^3}$
Bring the fractions at the same denominator:
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{A(x-2)^2}{(x-2)^3}+\dfrac{B(x-2)}{(x-2)^3}+\dfrac{C}{(x-2)^3}$
$x^2-6x+3=A(x-2)^2+B(x-2)+C$
$x^2-6x+3=A(x^2-4x+4)+Bx-2B+C$
$x^2-6x+3=Ax^2-4Ax+4A+Bx-2B+C$
$x^2-6x+3=Ax^2+(-4A+B)x+(4A-2B+C)$
Identify the coefficients and build the system:
$\begin{cases}
A=1\\
-4A+B=-6\\
4A-2B+C=3
\end{cases}$
Solve the system:
$\begin{cases}
A=1\\
-4(1)+B=-6\\
4(1)-2B+C=3
\end{cases}$
$\begin{cases}
A=1\\
-4+B=-6\\
4-2B+C=3
\end{cases}$
$\begin{cases}
A=1\\
B=-2\\
-2(-2)+C=-1
\end{cases}$
$\begin{cases}
A=1\\
B=-2\\
4+C=-1
\end{cases}$
$\begin{cases}
A=1\\
B=-2\\
C=-5
\end{cases}$
Write the decomposed fraction:
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
