Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$

Question

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?

Answer 1

$$\frac{x^3}{x^2+12x+36}=\frac{x^3}{(x+6)^2}=x+A+\frac B{x+6}+\frac C{(x+6)^2}$$

Answer 2

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.