Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed? 

$$ \begin{align}\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}&=
\dfrac{(x-3)(3(x-3)(x+2)^2-2(x+2)^3)}
{(x-3)^4}\\
&=           \dfrac{3(x-3)(x+2)^2-2(x+2)^3}{(x-3)^3}\\
&= \dfrac{(x+2)^2(2(x-3)-2(x+2))}{(x-3)^3} \\
&= \dfrac{(x+2)^2(3x-9-2x-4)}{(x-3)^3}\\
&= \dfrac{(x+2)^2(x-13)}{(x-3)^3}
\end{align}$$
