jmc

First, we simplify the denominator: $$\frac{2}{\sqrt[3]{4}+\sqrt[3]{32}}=$$$$\frac{2}{\sqrt[3]{4}+2\sqrt[3]{4}}=$$$$\frac{2}{3\sqrt[3]{4}}$$Then, we multiply the denominator by something that would remove the cube root. Multiplying $\sqrt[3]{4}$ by $\sqrt[3]{2}$ would give $\sqrt[3]{8}$, which is an integer, $2$. Therefore, we multiply the expression of $\frac{\sqrt[3]{2}}{\sqrt[3]{2}}$. $$\frac{2}{3\sqrt[3]{4}}\cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}}=$$$$\frac{2\sqrt[3]{2}}{6}=$$$$\frac{\sqrt[3]{2}}{3}$$Therefore, $A+B=2+3=\boxed{5}$.