Answer
$\dfrac{4u}{v^2}$
Work Step by Step
RECALL:
(i) $a^{-m} = \dfrac{1}{a^m},a\ne0$
(ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
(iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
Simplify the radicand by canceling common factors:
$\require{cancel}
=\sqrt{\dfrac{16\cancel{u^3}u^2\cancel{v}}{\cancel{u}\cancel{v^{5}v^4}}}
\\=\sqrt{\dfrac{16u^2}{v^4}}$
Factor the radicand so that at least one factor is a perfect square to obtain:
$\\=\sqrt{\dfrac{(4u)^2}{(v^2)^2}}$
Simplify to obtain:
$=\dfrac{4u}{v^2}$
