jmc

First, we calculate $(\sqrt{2}+\sqrt{3}{)}^2$: $$\begin{array}{rl}(\sqrt{2}+\sqrt{3}{)}^2& =(\sqrt{2}+\sqrt{3})(\sqrt{2}+\sqrt{3})\\ & =(\sqrt{2})(\sqrt{2})+(\sqrt{2})(\sqrt{3})+(\sqrt{3})(\sqrt{2})+(\sqrt{3})(\sqrt{3})\\ & =2+\sqrt{6}+\sqrt{6}+3\\ & =5+2\sqrt{6}.\end{array}$$ Multiplying this by $\sqrt{2}+\sqrt{3}$ gives $$\begin{array}{rl}(\sqrt{2}+\sqrt{3}{)}^3& =(\sqrt{2}+\sqrt{3}{)}^2(\sqrt{2}+\sqrt{3})\\ & =(5+2\sqrt{6})(\sqrt{2}+\sqrt{3})\\ & =5\sqrt{2}+5\sqrt{3}+(2\sqrt{6})(\sqrt{2})+(2\sqrt{6})(\sqrt{3})\\ & =5\sqrt{2}+5\sqrt{3}+2\sqrt{12}+2\sqrt{18}\\ & =5\sqrt{2}+5\sqrt{3}+2(2\sqrt{3})+2(3\sqrt{2})\\ & =11\sqrt{2}+9\sqrt{3}.\end{array}$$ Therefore, we have $a+b+c=\boxed{20}$. (Notice that $c=0;$ tricky!)

We may also expand $(\sqrt{3}+\sqrt{2}{)}^3$ using the Binomial Theorem, giving us ${\sqrt{2}}^3+3{\sqrt{2}}^2\sqrt{3}+3\sqrt{2}{\sqrt{3}}^2+{\sqrt{3}}^3$. Simplifying this yields $2\sqrt{2}+6\sqrt{3}+9\sqrt{2}+3\sqrt{3}=11\sqrt{2}+9\sqrt{3}$, and once again $a+b+c=\boxed{20}$.