If $\frac{x^2+5x+7}{(x-3{)}^3}=\frac{A}{(x-3)}+\frac{B}{(x-3{)}^2}+\frac{C}{(x-3{)}^3}$ , then $9A-3B+C=$

Given, $\frac{x^2+5x+7}{(x-3{)}^3}=\frac{A}{(x-3)}+\frac{B}{(x-3{)}^2}+\frac{C}{(x-3{)}^3}$

$\frac{x^2+5x+7}{(x-\cancel{3}{)}^3}=\frac{A(x-3{)}^2+B(x-3)+C}{(x-\cancel{3}{)}^2}$

$x^2+5x+7=A(x-3{)}^2+B(x-3)+C$

Put $x=3$

$$\begin{gathered} 9+15+7=A\left(0\right)+B\left(0\right)+C \\ \Rightarrow c=31 \end{gathered}$$

$x^2+5x+7=A(x^2-6x+4)+B(x-3)+C$

$x^2+5x+7=A{x}^2+(B-6)x+9A-3B+C$

Comparing $x_1, x^2$ coefficients & constant coefficients

$$\begin{gathered} A=1, B-6=5 \\ B=11 \end{gathered}$$

$\Rightarrow 9\left(1\right)-3\left(11\right)+31$

$\Rightarrow 9-33+31=7$.