Answer
$\dfrac{3y\sqrt[5]{4x^4y^2}}{2}$
Work Step by Step
Write 8 as $2^3$ to obtain:
$=\dfrac{3xy^2}{\sqrt[5]{2^3xy^3}}$
RECALL:
For any real number $a$,
$\sqrt[5]{a^5}= a$
Rationalize the denominator by multiplying $\sqrt[5]{2^2x^4y^2}$ to both the numerator and the denominator. Simplify using the rule above to obtain:
$\require{cancel}
=\dfrac{3xy^2 \cdot \sqrt[5]{2^2x^4y^2}}{\sqrt[5]{2^3xy^3} \cdot \sqrt[5]{2^2x^4y^2}}
\\=\dfrac{3xy^2\sqrt[5]{4x^4y^2}}{\sqrt[5]{2^5x^5y^5}}
\\=\dfrac{3xy^2\sqrt[5]{4x^4y^2}}{2xy}
\\=\dfrac{3\cancel{x}y^{\cancel{2}}\sqrt[5]{4x^4y^2}}{2\cancel{xy}}
\\=\dfrac{3y\sqrt[5]{4x^4y^2}}{2}$
