jmc

From the given equation, $$x^2+{\left(\frac{2x}{x-2}\right)}^2-45=0.$$Let $a=x$ and $b=\frac{2x}{x-2}.$ Then $a^2+b^2=45,$ or $$(a+b{)}^2-2ab-45=0.$$In other words, $${\left(x+\frac{2x}{x-2}\right)}^2-\frac{4x^2}{x-2}-45=0.$$We can write this as $${\left(\frac{x^2}{x-2}\right)}^2-\frac{4x^2}{x-2}-45=0.$$Let $y=\frac{x^2}{x-2}.$ Then $y^2-4y-45=0,$ which factors as $(y-9)(y+5)=0.$ So, $$\left(\frac{x^2}{x-2}-9\right)\left(\frac{x^2}{x-2}+5\right)=0.$$Then $(x^2-9x+18)(x^2+5x-10)=0,$ which factors as $$(x-3)(x-6)(x^2+5x-10)=0.$$If $x=3,$ then $$\frac{(x-2{)}^2(x+3)}{2x-3}=2.$$If $x=6,$ then $$\frac{(x-2{)}^2(x+3)}{2x-3}=16.$$If $x^2+5x-10=0,$ then $x^2=-5x+10,$ and $$x^3=x(-5x+10)=-5x^2+10x=-5(-5x+10)+10x=35x-50.$$Hence, $$\begin{array}{rl}\frac{x^3-x^2-8x+12}{2x-3}& =\frac{(35x-50)-(-5x+10)-8x+12}{2x-3}\\ & =\frac{32x-48}{2x-3}=16.\end{array}$$Thus, the possible values of $\frac{(x-2{)}^2(x+3)}{2x-3}$ are $\boxed{2,16}.$