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\[\sum _{ x=1 }^{ \infty } \frac { { 3x }^{ 2 }+12x+16 }{ { (x(x+1)(x+2)(x+3)(x+4)) }^{ 3 } } =\quad \frac { 1 }{ 4{ (a!) }^{ b } }\] Compute \(b-a\) if \(b\) and \(a\) are positive integers.

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