If frac{x-4}{x^2-5x+6} can be expanded in the | vedclass

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(A) First,we decompose the expression into partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$,we get

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$\frac{-73}{648}$

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(A) First,we decompose the expression into partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.Solving for $A$ and $B$,we get $\frac{2}{x-2} - \frac{1}{x-3}$.Rewrite the expression as: $2(x-2)^{-1} - (x-3)^{-1} = 2(-2)^{-1}(1-\frac{x}{2})^{-1} - (-3)^{-1}(1-\frac{x}{3})^{-1}$.$= -\left[1 + \frac{x}{2} + (\frac{x}{2})^2 + (\frac{x}{2})^3 + \dots\right] + \frac{1}{3}\left[1 + \frac{x}{3} + (\frac{x}{3})^2 + (\frac{x}{3})^3 + \dots\right]$.The coefficient of $x^3$ is $-\left(\frac{1}{2}\right)^3 + \frac{1}{3}\left(\frac{1}{3}\right)^3$.$= -\frac{1}{8} + \frac{1}{81} = \frac{-81 + 8}{648} = -\frac{73}{648}$.

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

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