Answer
The partial fraction is $\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$
Work Step by Step
$\frac{{{x}^{2}}-6x+3}{{{\left( x-2 \right)}^{3}}}=\frac{A}{\left( x-2 \right)}+\frac{B}{{{\left( x-2 \right)}^{2}}}+\frac{C}{{{\left( x-2 \right)}^{3}}}$
Now, multiply both sides by ${{\left( x-2 \right)}^{3}}$; then we get,
$\begin{align}
& {{x}^{2}}-6x+3=A{{\left( x-2 \right)}^{2}}+B\left( x-2 \right)+C \\
& =A\left( {{x}^{2}}+4-4x \right)+Bx-2B+C \\
& =A{{x}^{2}}+4A-4Ax+Bx-2B+C \\
& =A{{x}^{2}}+\left( -4A+B \right)x+4A-2B+C
\end{align}$
Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term:
$A=1$ …… (1)
$-4A+B=-6$ …… (2)
$4A-2B+C=3$ …… (3)
And put the value of A in equation (2):
$\begin{align}
& -4\left( 1 \right)+B=-6 \\
& B=-6+4 \\
& =-2
\end{align}$
…… (4)
Also, put the value of A and B in equation (3):
$\begin{align}
& 4\left( 1 \right)-2\left( -2 \right)+C=3 \\
& C=3-8 \\
& =-5
\end{align}$
Therefore,
$\frac{{{x}^{2}}-6x+3}{{{\left( x-2 \right)}^{3}}}=\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$
Thus, the partial fraction of the provided expression is $\frac{1}{\left( x-2 \right)}-\frac{2}{{{\left( x-2 \right)}^{2}}}-\frac{5}{{{\left( x-2 \right)}^{3}}}$.
