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? 
 A: $$ \begin{align}\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}&=
\dfrac{(x-3)(3(x-3)(x+2)^2-2(x+2)^3)}
{(x-3)^4}\\
&=           \dfrac{3(x-3)(x+2)^2-2(x+2)^3}{(x-3)^3}\\
&= \dfrac{(x+2)^2(2(x-3)-2(x+2))}{(x-3)^3} \\
&= \dfrac{(x+2)^2(3x-9-2x-4)}{(x-3)^3}\\
&= \dfrac{(x+2)^2(x-13)}{(x-3)^3}
\end{align}$$
A: Trusty maxima tells me:
factor(ratsimp((3*(x + 2)^2 * (x - 3)^2 - 2 * (x + 2)^3 * (x - 3))/(x - 3)^4));

is:
$$
\frac{(x - 13) (x + 2)^2}{(x - 3)^3}
$$
A trick that helps catch silly errors while simplifying is to replace some simple values, like $x = 0$ and $x = \pm 1$ in the expresssions.
A: $$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
Let's factor the $(x-3)$ in the numerator and denominator.
$$\dfrac{(x-3)\left[3(x+2)^2(x-3) - 2(x+2)^3\right]}{(x-3)(x-3)^3}$$
Now we can cancel out $(x-3)$ in the numerator and the denominator. That gives us:
$$\dfrac{3(x+2)^2(x-3) - 2(x+2)^3}{(x-3)^3}$$
Let us expand the equation.
$$\dfrac{3(x+2)^2(x-3) - 2(x+2)^3}{(x-3)^3}$$
$$=\dfrac{3(x^2+4x+4)(x-3)+2(x^3-6x^2+12x-8)}{x^3-9x^2+27x-27}$$
$$=\dfrac{3x^3-3x^2+4x^2-12x+4x-12+2x^3-12x^2+24x+16}{x^3-9x^2+27x-27}$$
Combine like terms:
$$\dfrac{3x^3-3x^2+4x^2-12x+4x-12+2x^3-12x^2+24x+16}{x^3-9x^2+27x-27}$$
$$=\dfrac{5x^3-11x^2-14x+4}{x^3-9x^2+27x-27}$$
We cannot factor the numerator nor simplify it any further. So, the answer to your question is:
$$\displaystyle \boxed{\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}=\dfrac{5x^3-11x^2-14x+4}{x^3-9x^2+27x-27}}$$
