Answer
$\color{blue}{\dfrac{h\sqrt[4]{9g^3hr^2}}{3r^2}}$
Work Step by Step
Rationalize the denominator by multiplying $9r^2$ to both the numerator and denominator of the radicand to obtain:
$=\sqrt[4]{\dfrac{g^3h^5(9r^2)}{9r^6(9r^2)}}
\\=\sqrt[4]{\dfrac{9g^3h^5r^2}{81r^8}}
\\=\sqrt[4]{\dfrac{9g^3h^5r^2}{3^4r^8}}
\\=\sqrt[4]{\dfrac{9g^3h^5r^2}{(3r^2)^4}}$
Bring out the fourth root of the denominator to obtain:
$\\=\dfrac{\sqrt[4]{9g^3h^5r^2}}{3r^2}$
Factor the radicand such that at least one factor is a perfect fourth power to obtain:
$\\=\dfrac{\sqrt[4]{h^4(9g^3hr^2)}}{3r^2}$
Bring out the fourth root of the perfect fourth power factor/s of the numerator to obtain:
$\\=\color{blue}{\dfrac{h\sqrt[4]{9g^3hr^2}}{3r^2}}$
