If frac{x^2+5x+7}{(x-3)^3}=frac{A}{(x-3)}+fra | vedclass

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(C) Given,$\frac{x^2+5x+7}{(x-3)^3} = \frac{A}{(x-3)} + \frac{B}{(x-3)^2} + \frac{C}{(x-3)^3}$ Multiply both sides by $(x-3)^3$: $x^2+5x+7 = A(x-3)^2

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$7$

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(C) Given,$\frac{x^2+5x+7}{(x-3)^3} = \frac{A}{(x-3)} + \frac{B}{(x-3)^2} + \frac{C}{(x-3)^3}$ Multiply both sides by $(x-3)^3$: $x^2+5x+7 = A(x-3)^2 + B(x-3) + C$ Let $u = x-3$,so $x = u+3$. Substitute $x = u+3$ into the equation: $(u+3)^2 + 5(u+3) + 7 = Au^2 + Bu + C$ $(u^2+6u+9) + 5u + 15 + 7 = Au^2 + Bu + C$ $u^2 + 11u + 31 = Au^2 + Bu + C$ Comparing coefficients: $A = 1, B = 11, C = 31$ Now calculate $9A - 3B + C$: $9(1) - 3(11) + 31 = 9 - 33 + 31 = 7$

If $\frac{x^2+5x+7}{(x-3)^3}=\frac{A}{(x-3)}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3}$,then $9A-3B+C=$

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