Answer
$\dfrac{5\sqrt[5]{2x^3}}{x}$
Work Step by Step
Write 16 as $2^4$ to obtain:
$=\dfrac{10}{\sqrt[5]{2^4x^2}}$
RECALL:
For any real number $a$,
$\sqrt[5]{a^5}= a$
Rationalize the denominator by multiplying $\sqrt[5]{2x^3}$ to both the numerator and the denominator. Simplify using the rule above to obtain:
$\require{cancel}
=\dfrac{10 \cdot \sqrt[5]{2x^3}}{\sqrt[5]{2^4x^2} \cdot \sqrt[5]{2x^3}}
\\=\dfrac{10\sqrt[5]{2x^3}}{\sqrt[5]{2^5x^5}}
\\=\dfrac{10\sqrt[5]{2x^3}}{2x}
\\=\dfrac{5\cancel{10}\sqrt[5]{2x^3}}{\cancel{2}x}
\\=\dfrac{5\sqrt[5]{2x^3}}{x}$
