Answer
The partial fraction decomposition is $\frac{1}{{{x}^{2}}-ax-bx+ab}=\frac{1/\left( a-b \right)}{x-a}+\frac{1/\left( b-a \right)}{x-b}$.
Work Step by Step
We know that the partial fraction can be written as,
$\begin{align}
& \frac{1}{{{x}^{2}}-ax-bx+ab}=\frac{1}{x\left( x-a \right)-b\left( x-a \right)} \\
& =\frac{1}{\left( x-a \right)\left( x-b \right)}
\end{align}$
And,
$\begin{align}
& \frac{1}{\left( x-a \right)\left( x-b \right)}=\frac{A}{x-a}+\frac{B}{x-b} \\
& =\frac{A\left( x-b \right)+B\left( x-a \right)}{\left( x-a \right)\left( x-b \right)}
\end{align}$
And,
$1=A\left( x-b \right)+B\left( x-a \right)$ …… (1)
Substitute the value of $x=a$ in equation (1),
$\begin{align}
& 1=A\left( a-b \right) \\
& A=\frac{1}{a-b}
\end{align}$
Again, put the value of $x=b$ in equation (1),
$\begin{align}
& 1=B\left( b-a \right) \\
& B=\frac{1}{b-a}
\end{align}$
Therefore,
$\frac{1}{{{x}^{2}}-ax-bx+ab}=\frac{1/\left( a-b \right)}{x-a}+\frac{1/\left( b-a \right)}{x-b}$
Thus, the partial fraction decomposition is $\frac{1}{{{x}^{2}}-ax-bx+ab}=\frac{1/\left( a-b \right)}{x-a}+\frac{1/\left( b-a \right)}{x-b}$.
