Answer
$\dfrac{3\sqrt[5]{4x^2}}{x}$
Work Step by Step
Write 8 as $2^3$ to obtain:
$=\dfrac{6}{\sqrt[5]{2^3x^3}}$
RECALL:
For any real number $a$,
$\sqrt[5]{a^5}= a$
Rationalize the denominator by multiplying $\sqrt[5]{2^2x^2}$ to both the numerator and the denominator. Simplify using the rule above to obtain:
$\require{cancel}
=\dfrac{6 \cdot \sqrt[5]{2^2x^2}}{\sqrt[5]{2^3x^3} \cdot \sqrt[5]{2^2x^2}}
\\=\dfrac{6\sqrt[5]{4x^2}}{\sqrt[5]{2^5x^5}}
\\=\dfrac{6\sqrt[5]{4x^2}}{2x}
\\=\dfrac{3\cancel{6}\sqrt[5]{4x^2}}{\cancel{2}x}
\\=\dfrac{3\sqrt[5]{4x^2}}{x}$
