solution: 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 $a + b + c = \boxed{20}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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 $a+b+c$?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 20
extra_info.question: 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 $a+b+c$?
extra_info.split: test