# How do you simplify sqrt(8u^8 v^3) sqrt(2u^5 v^6)?

Jul 2, 2015

$\sqrt{8 {u}^{8} {v}^{3}} \sqrt{2 {u}^{5} {v}^{6}} = 4 {u}^{6} {v}^{4} \sqrt{u v}$

with restriction $u , v \ge 0$

#### Explanation:

If $\sqrt{8 {u}^{8} {v}^{3}}$ is defined then $8 {u}^{8} {v}^{3} \ge 0$

If $\sqrt{2 {u}^{5} {v}^{6}}$ is defined then $2 {u}^{5} {v}^{6} \ge 0$

Use $\sqrt{a} \sqrt{b} = \sqrt{a b}$ when $a , b \ge 0$ to get:

$\sqrt{8 {u}^{8} {v}^{3}} \sqrt{2 {u}^{5} {v}^{6}}$

$= \sqrt{8 {u}^{8} {v}^{3} \cdot 2 {u}^{5} {v}^{6}}$

$= \sqrt{16 {u}^{13} {v}^{9}}$

$= \sqrt{{\left(4 {u}^{6} {v}^{4}\right)}^{2} u v}$

$= 4 {u}^{6} {v}^{4} \sqrt{u v}$