Answer
$1$
Work Step by Step
Factor what we can:
$ 4a^{2}+2ab+b^{2}= \qquad ...$ prime
$4a^{2}-b^{2}$ = difference of squares = $(2a-b)(2a+b)$
$8a^{3}-b^{3}$= difference of cubes = $(2a)^{3}-(b)^{3}$
$=(2a-b)[(2a)^{2}+(2a)(b)+(b)^{2}]$
$=(2a-b)(4a^{2}+2ab+b^{2})$
Rewrite the problem:
$\displaystyle \frac{(4a^{2}+2ab+b^{2})}{(2a+b)}\cdot\frac{(2a-b)(2a+b)}{(2a-b)(4a^{2}+2ab+b^{2})}=\qquad$ ... reduce common factors
$=\displaystyle \frac{(1)}{(2a+b)}\cdot\frac{(2a-b)(2a+b)}{(2a-b)(1)}$
$=\displaystyle \frac{(1)}{(2a+b)}\cdot\frac{(1)(2a+b)}{(1)(1)}$
$=\displaystyle \frac{(1)}{(1)}\cdot\frac{(1)(1)}{(1)(1)}$
= $1$
