I have that
$$\frac{x^2(1-x)}{(1-x)^3} = \frac{x^2}{(1-x)^3} -\frac{x^3}{(1-x)^3} $$
Assuming we begin by cancelling the $(1-x)$ on top with one of those on the bottom, how do we go about splitting $\dfrac{x^2}{(1-x)^{2}}$ into the two partial fractions shown (previously, I have only dealt with partial fractions where the denominator power is greater than the numerator). All help appreciated!
Actually, the first step should be$$\frac{x^2(1-x)}{(1-x)^3}=\frac{x^2}{(1-x)^2}.$$And now note that\begin{align}\frac{x^2}{(1-x)^2}&=\frac{\bigl((x-1)+1\bigr)^2}{(x-1)^2}\\&=\frac{(x-1)^2+2(x-1)+1}{(x-1)^2}\\&=1+\frac{2(x-1)+1}{(x-1)^2}\\&=1+\frac2{x-1}+\frac1{(x-1)^2}.\end{align}
$$\displaylines{{x^2(1-x)\over (1-x)^3}={x^2\over (x-1)^2}={x^2-1\over (x-1)^2} +{1\over (x-1)^2} \\ = {(x-1)+2\over x-1}+{1\over (x-1)^2}=1+{2\over x-1}+{1\over (x-1)^2}}$$
