Let
\[\sum_{n = 0}^{123456789} \frac{3n^2 + 9n + 7}{(n^2 + 3n + 2)^3} = \frac{a}{b},\]where $a$ and $b$ are relatively prime positive integers.  Find $b - a.$
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<|im_start|>user
Let
\[\sum_{n = 0}^{123456789} \frac{3n^2 + 9n + 7}{(n^2 + 3n + 2)^3} = \frac{a}{b},\]where $a$ and $b$ are relatively prime positive integers.  Find $b - a.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
1
Let
\[\sum_{n = 0}^{123456789} \frac{3n^2 + 9n + 7}{(n^2 + 3n + 2)^3} = \frac{a}{b},\]where $a$ and $b$ are relatively prime positive integers.  Find $b - a.$
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