Use the Quotient Property to simplify square roots.

(a) $\sqrt{\frac{27p^{2}q}{108p^4{q}^3}}$

(b)$\sqrt[3]{\frac{16c^5{d}^7}{250c^2{d}^2}}$

(c)$\sqrt[6]{\frac{2m^9{n}^7}{128m^{3}n}}$

Given expression:

$\sqrt{\frac{27p^{2}q}{108p^4{q}^3}}$.

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number, $b\ne 0$for any integer $n\ge 2$.

$$\begin{gathered} \sqrt{\frac{27p^{2}q}{108p^4{q}^3}}=\sqrt{\frac{27}{108}.\frac{1}{p^{4-2}}.\frac{1}{q^{3-1}}} \\ =\sqrt{\frac{9}{36}.\frac{1}{p^2}.\frac{1}{q^2}} \\ =\sqrt{{\left(\frac{3}{6}\right)}^2.\frac{1}{p^2}.\frac{1}{q^2}} \\ =\frac{1}{2\left|pq\right|} \end{gathered}$$

Given radical expression:

$\sqrt[3]{\frac{16c^5{d}^7}{250c^2{d}^2}}$.

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number, $b\ne 0$for any integer $n\ge 2$.

$$\begin{gathered} \sqrt[3]{\frac{16c^5{d}^7}{250c^2{d}^2}}=\sqrt[3]{\frac{16}{250}·c^{5-2}·d^{7-2}} \\ =\sqrt[3]{\frac{8}{125}·c^3·d^5} \\ =\sqrt[3]{{\left(\frac{2}{5}\right)}^3·c^3·d^3·d^2} \\ =\frac{2cd\sqrt[3]{d^2}}{5} \end{gathered}$$

Given expression:

$\sqrt[3]{\frac{16c^5{d}^7}{250c^2{d}^2}}$.

The radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number, $b\ne 0$for any integer $n\ge 2$.

$$\begin{gathered} \sqrt[6]{\frac{2m^9{n}^7}{128m^{3}n}}=\sqrt[6]{\frac{2}{128}·m^{9-3}·n^{7-1}} \\ =\sqrt[6]{\frac{1}{64}·m^6·n^6} \\ =\sqrt[6]{\frac{1}{2^6}}·\sqrt[6]{m^6}·\sqrt[6]{n^6} \\ =\frac{1}{2}\left|mn\right| \end{gathered}$$