Is the below the correct partial fraction decomposition?
$$\frac{s^2 - 6s + 9}{(s-2)^3}=\frac{A}{s-2}+\frac{B}{(s-2)^2}+\frac{C}{(s-2)^3}$$
I can see that the numerator doesn't have a factor of $(s-2)$ so I can't see how I can simplify it any further. But I wasn't sure if that is the proper method for decomposing the factors on the next denominators.
Thank you.
Just note that the numerator is $(s-3)^2$, then you can use
$(\color{red}{x}-\color{green}{y})^2=\color{red}{x^2}-2\color{red}{x}\color{green}{y}+\color{green}{y^2}$
which gives
$${(\color{red}{s-2}-\color{green}{1})^2\over (s-2)^3}={\color{red}{(s-2)^2}-2\color{red}{(s-2)}\cdot\color{green}{1}+\color{green}{1^2}\over (s-2)^3}$$
this gives
$$A=1,\; B=-2,\; C=1$$
