solution: The sum of three consecutive integers takes the form $(k-1)+(k)+(k+1)=3k$ and hence is a multiple of 3.  Conversely, if a number $n$ is a multiple of 3, then $n/3-1$, $n/3$, and $n/3+1$ are three consecutive integers that sum to give $n$.  Therefore, a number is a sum of three consecutive integers if and only if it is a multiple of 3.  The smallest positive perfect cube that is a multiple of 3 is $3^3=\boxed{27}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  What is the smallest positive perfect cube that can be written as the sum of three consecutive integers?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 27
extra_info.question: What is the smallest positive perfect cube that can be written as the sum of three consecutive integers?
extra_info.split: test