If $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$,then:

  • A
    $a = 2$
  • B
    $b = 1$
  • C
    $a = 3$
  • D
    $b = 4$

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