Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$ Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. So how would you use partial fraction decomposition? 

After the observation that the denominator is equal to $(x+6)^2$ (which was mentioned in other answer), this problem can be solved relatively easy using the substitution $y=x+6$.
$$
\begin{align*}
\frac{x^3}{(x+6)^2}
&=\frac{(y-6)^3}{y^2}=\\
&=\frac{y^3-3\cdot 6 y^2+3\cdot 6^2 y - 6^3}{y^2}=\\
&=\frac{y^3-18y^2+108y-216}{y^2}=\\
&=y-18+\frac{108}y-\frac{216}{y^2}=\\
&=x-12+\frac{108}{x+6}-\frac{216}{(x+6)^2}
\end{align*}$$
You can check the result using WolframAlpha.
