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