Answer
$3x^2y^{3} \sqrt[4]{2x}$.
Work Step by Step
Factor the radicand so that some factors are fourth powers:
$\sqrt[4]{162x^9y^{12}}\\
=\sqrt[4]{(81\cdot 2)(x^8\cdot x)y^{12}}$
$=\sqrt[4]{81x^8y^{12}\cdot 2x}$
With $81=3^4,x^8=(x^2)^4$ and $y^{12}=(y^3)^4$, use the rule $a^nb^nc^n=(abc)^n$. to obtain:
$=\sqrt[4]{3^4(x^2)^4(y^{3})^{4}\cdot 2x}\\
=\sqrt[4]{(3x^2y^3)^{4}\cdot 2x}$
Separate using the rule $\sqrt[n]{a\cdot b}=\sqrt[n]{a} \cdot \sqrt[n]{b}, a,b>0$ to obtain:.
$=\sqrt[4]{(3x^2y^3)^{4}}\cdot \sqrt[4]{2x}$
Simplify.
$=3x^2y^{3} \sqrt[4]{2x}$
Hence, the correct answer is $3x^2y^{3}\sqrt[4]{2x}$.
