You are making this so much more difficult on yourself than you have to.

Notice that $\displaystyle \begin{align*} \frac{1}{\left( 1 - 2\,x \right) ^3} = \left[ 1 + \left( -2\,x \right) \right] ^{-3} \end{align*}$, which can be expanded using the Binomial Series $\displaystyle \begin{align*} \left( 1 + X \right) ^{\alpha} = \sum_{k = 0}^{\infty}{ {\alpha\choose{k}}\,X^k }= 1 + \alpha\,X + \frac{\alpha \,\left( \alpha - 1 \right) }{2!}\,X^2 + \frac{\alpha \, \left( \alpha - 1 \right) \, \left( \alpha - 2 \right) }{3!}\,X^3 + \dots \end{align*}$

Once you have that, multiply through by $\displaystyle \begin{align*} \left( x - x^2 \right) \end{align*}$.

Notice that as $\displaystyle \begin{align*} \alpha < -1 \end{align*}$ this will only be convergent where $\displaystyle \begin{align*} \left| X \right| < 1 \end{align*}$, i.e. where $\displaystyle \begin{align*} \left| x \right| < \frac{1}{2} \end{align*}$.