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? 
 A: $$\frac{x^3}{x^2+12x+36}=\frac{x^3}{(x+6)^2}=x+A+\frac B{x+6}+\frac C{(x+6)^2}$$
A: 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.
A: \begin{array}{rcrrrrrrr}
           &    &  x & -12 \\
           &    & -- & --- &    --- &   --- & \\
x^2+12x+36 & )  & x^3 \\
           &    & x^3 &  12x^2 &   36x \\
           &    &  -- &    --- &   --- \\
           &    &     & -12x^2 &  -36x \\ 
           &    &     & -12x^2 & -144x & -432 \\
           &    &     &    --- &   --- &  --- \\
           &    &     &        &  108x &  432 \\
\end{array}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108x+432}{(x+6)^2}$$
where 
\begin{align}
   \dfrac{108x+432}{(x+6)^2} &= \dfrac{A}{x+6} + \dfrac{B}{(x+6)^2} \\
   108x + 432 &= A(x+6) + B
\end{align}
$$ \text{Let $x=-6$ and you get $B = -216$.}$$
$$ \text{Let $B = -216$ and you get}$$
\begin{align}
   108x + 432 &= A(x+6) - 216 \\
   A(x+6) &= 108x + 648 \\
   A &= 108
\end{align}
And so
$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108}{x+6} - \dfrac{216}{(x+6)^2}$$
