Show that $\displaystyle \frac{2^\frac{2}{3}}{4}=\sqrt[3]{\frac{1}{16}}$.
Any help showing detailed working out would be appreciated!
$\displaystyle \frac{2^\frac{2}{3}}{4}=\sqrt[3]{\frac{1}{16}}$
$\displaystyle \displaystyle\frac{2^\frac{2}{3}}{4}$
$\displaystyle =\displaystyle\frac{2^\frac{2}{3}}{2^2}$
$\displaystyle =\displaystyle(2^\frac{2}{3})(2^{-2})$
Using the law of indices, $\displaystyle (x^a)(x^b)=(x^{a+b})$
Apply that here, then tidy up to give you the answer you seek.
Edit: Unfortunately, I am far slower than Plato, as usual. =.=