<s>[INST] Finding a power series representation for $\left(\frac{x}{2-x}\right)^3$ 
Find a power series representation for $\displaystyle\left(\frac{x}{2-x}\right)^3$

My approach is in finding something similar to $\displaystyle\left(\frac{x}{2-x}\right)^3$ to which I can easily find the power series representation of. 
I use $\displaystyle\frac{1}{2-x}$, noting that $\displaystyle\left(\frac{1}{2-x}\right)'=\frac{1}{(2-x)^2} \text { and } \left(\frac{1}{2-x}\right)''=\frac{2}{(2-x)^3}$.
So 
$$\displaystyle \frac{1}{2-x}=\int\frac{1}{(2-x)^2}dx \iff \frac{1}{2}\sum^{\infty}_{n=0}\left(\frac{x}{2}\right)^n=\int\frac{1}{(2-x)^2}dx$$
and differentiating both sides, I get the power series representation of the first derivative
$$\frac{1}{2}\sum^{\infty}_{n=1}n\left(\frac{x}{2}\right)^{n-1}=\frac{1}{(2-x)^2}$$
for the second derivative, 
$$\frac{1}{(2-x)^2}=\int\frac{2}{(2-x)^3} \iff \frac{1}{2}\sum^{\infty}_{n=1}n\left(\frac{x}{2}\right)^{n-1}=\frac{1}{2}\sum^{\infty}_{n=0}(n+1)\left(\frac{x}{2}\right)^n=\int\frac{2}{(2-x)^3}dx$$
differentiating both sides, I get the power series representation of the second derivative
$$\displaystyle \frac{1}{2}\sum^{\infty}_{n=1}(n+1)n\left(\frac{x}{2}\right)^{n-1}=\frac{2}{(2-x)^3}$$
Is this so far correct? If it is, in the end I would multiply the power series representation of $\displaystyle\frac{2}{(2-x)^3}$ by $\displaystyle\frac{x^3}{2}$ to cancel out the $2$ and get the power series for $\displaystyle\left(\frac{x}{2-x}\right)^3$.
 [/INST] You're approach is fine (besides the missing integration constant according to @Winther's comment).

Hint: Note, that you could also obtain a power series representation by applying the binomial series to your expression, since
\begin{align*}
\left(\frac{x}{2-x}\right)^3
=\left(\frac{x}{2}\right)^3\left(1-\frac{x}{2}\right)^{-3}
=\ldots
\end{align*}

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