Answer
$\frac{c+d}{2}$
Work Step by Step
Factor and simplify, we have:
$\frac{ac+ad+bc+bd}{a^2-b^2}\cdot\frac{a^3-b^3}{2a^2+2ab+2b^2}=\frac{a(c+d)+b(c+d)}{(a+b)(a-b)}\cdot\frac{(a-b)(a^2+ab+b^2)}{2(a^2+ab+b^2)}=\frac{(c+d)(a+b)}{(a+b)}\cdot\frac{1}{2}=\frac{c+d}{2}$ where $a\pm b\ne0, a^2+ab+b^2\ne0$
