+118608
I am going to assume that you really mean
(x-2)^2/(4y)*16y^3/(4-x^2)
$$\\\dfrac{\frac{(x-2)^2}{4y}*16y^3}{(4-x^2)}}\\\\\\
=\dfrac{\frac{(x-2)^2}{1}*4y^2}{(4-x^2)}}\\\\\\
=\frac{(x-2)^2*4y^2}{(2-x)(2+x)}}\\\\\\
=\frac{(x-2)*4y^2}{(2+x)}}\\\\
=\frac{4y^2(x-2)}{2+x}}\\\\
or\\\\
=\frac{4y^2x-8y^2}{2+x}}\\\\$$
+118608
I am going to assume that you really mean
(x-2)^2/(4y)*16y^3/(4-x^2)
$$\\\dfrac{\frac{(x-2)^2}{4y}*16y^3}{(4-x^2)}}\\\\\\
=\dfrac{\frac{(x-2)^2}{1}*4y^2}{(4-x^2)}}\\\\\\
=\frac{(x-2)^2*4y^2}{(2-x)(2+x)}}\\\\\\
=\frac{(x-2)*4y^2}{(2+x)}}\\\\
=\frac{4y^2(x-2)}{2+x}}\\\\
or\\\\
=\frac{4y^2x-8y^2}{2+x}}\\\\$$
