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} $.