solution: We first cube each side of the equation to get $x^2 - 4x + 4 = 16^3$.  Notice that $x^2 - 4x + 4 = (x-2)^2.$
  
  Therefore, we have that $x-2 = \pm 16^{3/2} = \pm 64$.  Therefore, the possible values of $x$ are $-62$ and $66,$ and the only positive value is therefore $\boxed{66}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Solve for the positive value of $x$ such that $\sqrt[3]{x^2 - 4x + 4} = 16$.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 66
extra_info.question: Solve for the positive value of $x$ such that $\sqrt[3]{x^2 - 4x + 4} = 16$.
extra_info.split: test