The number $(\sqrt{2}+\sqrt{3})^3$ can be written in the form $a\sqrt{2} + b\sqrt{3} + c\sqrt{6}$, where $a$, $b$, and $c$ are integers.  What is $a+b+c$?
First, we calculate $(\sqrt{2} + \sqrt{3})^2$: \begin{align*}
(\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{align*} Multiplying this by $\sqrt{2} +\sqrt{3}$ gives \begin{align*}
(\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{align*} 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 $a + b + c = \boxed{20}$.