<s>[INST] Power Series Representation of $f(x) = \frac{x^3}{(2-x)^3}$ I am trying to represent the function, $f(x) = \frac{x^3}{(2-x)^3}$ as a power series and saw this on a different post. 
$
{1\over (1-u)} =\sum_{n=0}^{\infty}u^{n}, \qquad|u|<1, \tag1
$
$-{1 \over (1-u)^2} =\sum_{n=1}^{\infty}nu^{n-1}, \qquad|u|<1,\tag2$
${2 \over (1-u)^3} =\sum_{n=2}^{\infty}n(n-1)u^{n-2}, \qquad|u|<1. \tag3$
I understand up to this part. I began with the general form 
$f(x) = \frac{1}{(1-x)}$
and derived it twice to get: 
$f''(x) = \frac{2}{(1-x)^3}$
which somewhat resembles the function I need to represent as a power series. However, this is as far as I got and keep hitting a dead-end. 
1) $f(x)=\frac{x^3}{(2-x)^3} = [x(\frac{1}{2-x})]^3
= [ \frac{x}{2}(\frac{1}{1-x/2})]  ^3,
$
2) $\frac{x^3}{2^3} (\frac{1}{1-x/2})^3  
= \frac{x^3}{2^3}\sum_{n=2}^{\infty}n(n-1)(x/2)^{n-2}, 
$
3)Therefore,  $f(x) = \frac{x^3}{(2-x)^3}$
represented as a power series is: 
$\sum_{n=2}^{\infty}n(n-1)(x/2)^{n+1}, \qquad|x|<2
$
I have no solution sheet for this but I don't know if I am right or wrong. I also have a different answer from what I saw other people post online so if I am wrong, please tell me why. 
 [/INST] $$f(x)=\frac{x^3}{(2-x)^3}=\frac{x^3}{8}(1-\frac{x}{2})^{-3}=\frac{x^3}{8}[1-{-3 \choose 1}\frac{x}{2}+{-3 \choose 2} (\frac{x}{2})^2 -{-3 \choose 3}(\frac{x}{2})^3$$ $$+{-3 \choose k} (\frac{x}{2})^4 .....] ~~~~(1)$$
Using $${-p \choose k}=\frac{-p(-p-1)(-p-2)....(-p-k+1)}{k!},$$
we find that $${-3 \choose 1}=-3, {-3 \choose 2}=-3.-4/2=6, {-3 \choose 3}=-3.-4.-5/6 =-10, {-3\choose 4} =-3.-4.-5.-6/24 = 15....$$ So on and so forth.
Inserting these co-efficients in (1) we gwt a serie which is valid for $|x|<2$.
When $|x|>1$, then $$f(x)=-(1-\frac{2}{x})^{-3}=-\left [1-{-3\choose 1}(\frac{2}{x})+{-3 \choose 2} (\frac{2}{x})^2-{-3 \choose 3} (\frac{2}{x})^3+..\right] $$
</s>