Answer
$2x{{y}^{3}}\sqrt[4]{3{{x}^{3}}}$
Work Step by Step
Formula used:
Product rule to simplify expression:
$\sqrt[n]{ab}=\sqrt[n]{a}.\sqrt[n]{b}$
$\left( \sqrt[n]{a}\text{ and }\sqrt[n]{b} \right)$ must both be real numbers.
Consider expression $\sqrt[4]{48{{x}^{7}}{{y}^{12}}}$,
Identifying the largest perfect fourth power factor,
$\sqrt[4]{48{{x}^{7}}{{y}^{12}}}=\sqrt[4]{16\times {{x}^{4}}\times {{y}^{12}}\times 3\times {{x}^{3}}}$
Factoring into radicals,
$\sqrt[4]{16\times {{x}^{4}}\times {{y}^{12}}\times 3\times {{x}^{3}}}=\sqrt[4]{16\times {{x}^{4}}\times {{y}^{12}}}\times \sqrt[4]{3\times {{x}^{3}}}$
Find the fourth root. we assume $x\ge 0$,
$2x{{y}^{3}}\sqrt[4]{3{{x}^{3}}}$
Partial check:
$\begin{matrix}
{{\left( 2x{{y}^{3}}\sqrt[4]{3{{x}^{3}}} \right)}^{4}}\overset{?}{\mathop{=}}\,{{\left( 2 \right)}^{4}}{{\left( x \right)}^{4}}{{\left( {{y}^{3}} \right)}^{4}}{{\left( \sqrt[4]{3{{x}^{3}}} \right)}^{4}} \\
\overset{?}{\mathop{=}}\,16\times {{x}^{4}}\times {{y}^{12}}\times 3{{x}^{3}} \\
=48{{x}^{7}}{{y}^{12}} \\
\end{matrix}$
The expression $\sqrt[4]{48{{x}^{7}}{{y}^{12}}}$ can be simplified as $2x{{y}^{3}}\sqrt[4]{3{{x}^{3}}}$.
