jmc

We rationalize each of the fractions in parentheses, by using the sum and difference of cubes identities. First, we have $$\begin{array}{rl}\frac{5}{\sqrt[3]{3}+\sqrt[3]{2}}& =\frac{5\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}\\ & =\frac{5\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}{3+2}\\ & =\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}.\end{array}$$Similarly, $$\begin{array}{rl}\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}& =\frac{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}}{\left(\sqrt[3]{3}-\sqrt[3]{2}\right)\left(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\right)}\\ & =\frac{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}}{3-2}\\ & =\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}.\end{array}$$Therefore,$$\begin{array}{rl}\frac{1}{2}\left(\frac{5}{\sqrt[3]{3}+\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}\right)& =\frac{1}{2}\left(\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)+\left(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}\right)\right)\\ & =\sqrt[3]{9}+\sqrt[3]{4},\end{array}$$so $a+b=9+4=\boxed{13}.$