(x-3 / 2)^3 * (1 / x - 3)^2
Hmm.. You miss that $\displaystyle {\left( \frac{x-3}{2} \right ) }^3 = \frac{(x-3)^3}{8}$
It has to be, $\displaystyle {\left( \frac{x-3}{2} \right ) }^3 {\left ( \frac{1}{x-3} \right ) }^2$
$\displaystyle \frac{(x-3)^3}{8} \frac{1}{(x-3)^2}$
$\displaystyle \frac{x-3}{8}$