i am to use $\displaystyle \frac{1}{1-x}=\sum{x^n}=1+x+x^2+...$ provided $\displaystyle |x|<1$ to find a power series representation for $\displaystyle f(x)=\frac{x^3}{(x-2)^2}$

so i want to foil the binomial, move the cube, and convert the statement into the form of $\displaystyle \frac{1}{1-x}$

considering only $\displaystyle \frac{1}{(x-2)^2}$

i get $\displaystyle \frac{1}{x^2-4x+4}=\frac{1}{4-(4x-x^2)}=\frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}$

so $\displaystyle \frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}=\frac{1}{4}*\sum{(x-\frac{x^2}{4})^n}$

now i'm at a loss has to how to get it into the form $\displaystyle \sum{c_n*x^n}$

integrating/differentiating term-by-term doesn't seem to be helpful as each makes the situation far more complicated.