HELP!

cube root 16 =  2cube root 2

So the expression simplifies to    1 / [3 * ³(2)]......multiply top/bottom by  ³(2^2) and we  have

³(2^2) / 3 * ³(2^3)  = 

³(2^2) / 3* 2  =

³ [ 4 ] / 6

We can write $$\begin{align*} \frac{1}{\sqrt[3]{2}+\sqrt[3]{16}} &= \frac{1}{\sqrt[3]{2}+\sqrt[3]{4^2}}\ &= \frac{1}{\sqrt[3]{2}+2}\ &= \frac{\sqrt[3]{2}-2}{2(\sqrt[3]{2}-2)}\ &= \boxed{\frac{1}{2}}. \end{align*}$$