jmc

First, simplify $\sqrt[3]{16}=\sqrt[3]{2^3\cdot 2}=2\sqrt[3]{2}$. Substituting this, the fraction becomes: $$\frac{1}{\sqrt[3]{2}+2\sqrt[3]{2}}=\frac{1}{3\sqrt[3]{2}}$$ To rationalize this, we need to multiply the numerator and denominator by something that will eliminate the cube root in the denominator. If we multiply $\sqrt[3]{2}$, by $\sqrt[3]{4}$, then the result will be $\sqrt[3]{2}\cdot \sqrt[3]{4}=\sqrt[3]{2\cdot 4}=\sqrt[3]{8}=2$. So, multiply the expression above by $\frac{\sqrt[3]{4}}{\sqrt[3]{4}}$. $$\frac{1}{3\sqrt[3]{2}}\cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}}=\frac{\sqrt[3]{4}}{3\sqrt[3]{8}}=\boxed{\frac{\sqrt[3]{4}}{6}}$$