## Intermediate Algebra (6th Edition)

$\sqrt[6] x$
$\frac{\sqrt x}{\sqrt[3] x}=\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}}$ To simplify, we can use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers). Therefore, $\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}}=x^{\frac{3}{6}-\frac{2}{6}}=x^{\frac{1}{6}}=\sqrt[6] x$