solution: We know that
  \[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} > a,\]and
  \[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}} < a + 1.\]The integer $a$ that satisfies $a < \sqrt[3]{2} < a + 1$ is $a = 1.$
  
  Then
  \[\sqrt[3]{2} - 1 = \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}},\]so
  \[\frac{1}{\sqrt[3]{2} - 1} = b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}.\]As before, $b$ must satisfy
  \[b < \frac{1}{\sqrt[3]{2} - 1} < b + 1.\]Rationalizing the denominator, we get
  \[\frac{1}{\sqrt[3]{2} - 1} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{(\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1)} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{2 - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1.\]We have that
  \[\sqrt[3]{4} + \sqrt[3]{2} + 1 > 1 + 1 + 1 = 3.\]Also, $1.3^3 = 2.197 > 2$ and $1.6^3 = 4.096 > 4,$ so
  \[\sqrt[3]{4} + \sqrt[3]{2} + 1 < 1.3 + 1.6 + 1 = 3.9 < 4,\]so $b = \boxed{3}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  If
  \[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}},\]where $a,$ $b,$ $c,$ $d$ are positive integers, compute $b.$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 3
extra_info.question: If
  \[\sqrt[3]{2} = a + \cfrac{1}{b + \cfrac{1}{c + \cfrac{1}{d + \dotsb}}},\]where $a,$ $b,$ $c,$ $d$ are positive integers, compute $b.$
extra_info.split: test