If $\frac{x - 4}{x^{2} - 5 x + 6}$ can be expanded in the ascending power of x, then the coefficient of x³ is

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$\frac{- 73}{648}$

$\frac{73}{648}$

$\frac{- 71}{648}$

$\frac{71}{648}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} \frac{x - 4}{x^{2} - 5 x + 6} & = & \frac{A}{x - 2} + \frac{B}{x - 3} \\ x - 4 & = & A (x - 3) + B (x - 2) \\ x & = & 2 \Rightarrow - 2 = - A \Rightarrow A = 2 \\ x & = & 3 \Rightarrow - 1 = B \\ \frac{x - 4}{x^{2} - 5 x + 6} & = & \frac{2}{x - 2} = \frac{1}{x - 3} = \frac{- 1}{1 - \frac{x}{2}} + \frac{1}{3 (1 - \frac{x}{3})} \\ \Rightarrow & = - {(1 - \frac{x}{2})}^{- 1} + \frac{1}{3} {(1 - \frac{x}{3})}^{- 1} \\ = & - (1 + \frac{x}{2} + \frac{x^{2}}{4} + \frac{x^{3}}{8} + \dots \dots) \\ + \frac{1}{3} (1 + \frac{x}{3} + \frac{x^{2}}{9} + \frac{x^{3}}{27} + \dots \dots) \\ x^{3} coeff & = & \frac{- 1}{8} + \frac{1}{81} = - \frac{73}{648} \end{array}$