If $\frac{x-4}{x^2-5 x+6}$ can be expanded in the ascending powers of $x$, then the coefficient of $x^3$ is

If $\frac{x-4}{x^2-5 x+6}$ can be expanded in the ascending powers of $x$, then the coefficient of $x^3$ is
  1. $\frac{-73}{648}$
  2. $\frac{73}{648}$
  3. $\frac{71}{648}$
  4. $\frac{-71}{648}$

Solution

$\begin{aligned} & \frac{x-4}{x^2-5 x+6}=\frac{x-4}{(x-2)(x-3)} \\ & =\frac{2}{(x-2)}-\frac{1}{(x-3)}=2(x-2)^{-1}-(x-3)^{-1} \\ & =2(-2)^{-1}\left(1-\frac{x}{2}\right)^{-1}-(-3)^{-1}\left(1-\frac{x}{3}\right)^{-1} \\ & =-\left[1+\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3+\ldots\right] \\ & \quad+\frac{1}{3}\left[1+\left(\frac{x}{3}\right)+\left(\frac{x}{3}\right)^2+\left(\frac{x}{3}\right)^3+\ldots\right] \\ & \therefore \text { Coefficient of } x^3 \text { in } \frac{x-4}{x^2-5 x+6} .\end{aligned}$ $\begin{aligned} & =-\left(\frac{1}{2}\right)^3+\frac{1}{3}\left(\frac{1}{3}\right)^3 \\ & =-\frac{1}{8}+\frac{1}{81}=\frac{-81+8}{648} \\ & =-\frac{73}{648}\end{aligned}$

Asked in: AP EAMCET 2004

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