+118696
I am going to assume that you really mean
(x-2)^2/(4y)*16y^3/(4-x^2)
\\\dfrac{\frac{(x-2)^2}{4y}*16y^3}{(4-x^2)}}\\\\\\ =\dfrac{\frac{(x-2)^2}{1}*4y^2}{(4-x^2)}}\\\\\\ =\frac{(x-2)^2*4y^2}{(2-x)(2+x)}}\\\\\\ =\frac{(x-2)*4y^2}{(2+x)}}\\\\ =\frac{4y^2(x-2)}{2+x}}\\\\ or\\\\ =\frac{4y^2x-8y^2}{2+x}}\\\\
+118696
I am going to assume that you really mean
(x-2)^2/(4y)*16y^3/(4-x^2)
\\\dfrac{\frac{(x-2)^2}{4y}*16y^3}{(4-x^2)}}\\\\\\ =\dfrac{\frac{(x-2)^2}{1}*4y^2}{(4-x^2)}}\\\\\\ =\frac{(x-2)^2*4y^2}{(2-x)(2+x)}}\\\\\\ =\frac{(x-2)*4y^2}{(2+x)}}\\\\ =\frac{4y^2(x-2)}{2+x}}\\\\ or\\\\ =\frac{4y^2x-8y^2}{2+x}}\\\\
