Answer
$\dfrac{2x(x^2+4x+16)}{x+4} $
Work Step by Step
By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have
$$
\frac{4x^2}{x^2-16}\cdot \frac{x^3-64}{2x} = \frac{4x^2}{(x-4)(x+4)}\cdot \frac{(x-4)(x^2+4x+16)}{2x} \\
= \frac{2x(x^2+4x+16)}{x+4}
.$$
