Answer
$\dfrac{7\sqrt[3]{4x}}{2x}$
Work Step by Step
RECALL:
For any real number $a$,
$\sqrt[3]{a^3}= a$
Rationalize the denominator by multiplying $\sqrt[3]{2^2x}$ to both the numerator and the denominator. Simplify using the rule above to obtain:
$\require{cancel}
=\dfrac{7 \cdot \sqrt[3]{2^2x}}{\sqrt[3]{2x^2} \cdot \sqrt[3]{2^2x}}
\\=\dfrac{7\sqrt[3]{4x}}{\sqrt[3]{2^3x^3}}
\\=\dfrac{7\sqrt[3]{4x}}{2x}$
