Answer
$2x^{\frac{1}{6}}$
Work Step by Step
RECALL:
(i) $a^{-m} = \dfrac{1}{a^m},a\ne0$
(ii) $\sqrt[n]{a^m} = a^{\frac{m}{n}}$
(iii) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
Simplify the numerator to obtain:
$=\dfrac{\sqrt[3]{2^3x^2}}{\sqrt{x}}
\\=\dfrac{2\sqrt[3]{x^2}}{\sqrt{x}}$
Use rule (ii) above to obtain:
$=\dfrac{2x^{\frac{2}{3}}}{x^{\frac{1}{2}}}$
Use rule (iii) above to to obtain:
$=2x^{\frac{2}{3} - \frac{1}{2}}
\\=2x^{\frac{4}{6}-\frac{3}{6}}
\\=2x^{\frac{1}{6}}$
