Answer
$3x^2y^2\sqrt[3]{3x^2}$
Work Step by Step
Write 81 as $3^4$ to obtain:
$=\sqrt[3]{3^4x^8y^6}$
Factor the radicand (expression inside the radical sign) so that at least one factor is a perfect cube, and then simplify to obtain:
$=\sqrt[3]{(3^3x^6y^6)\cdot 3x^2}
\\=\sqrt[3]{3^3(x^2)^3(y^2)^3 \cdot 3x^2}
\\=3x^2y^2\sqrt[3]{3x^2}$
