begin{aligned} & frac{x^2+x+1}{(x-1)(x-2)(x-3 | vedclass

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(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ Multiplying both sides by $(x-1)(x-2)(x-3)$,we get: $x^2+x+1 = A

Vedclass · Accepted Answer

$8$

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(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ Multiplying both sides by $(x-1)(x-2)(x-3)$,we get: $x^2+x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$ To find $A$,put $x=1$: $1^2+1+1 = A(1-2)(1-3) \Rightarrow 3 = A(-1)(-2) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$ To find $C$,put $x=3$: $3^2+3+1 = C(3-1)(3-2) \Rightarrow 13 = C(2)(1) \Rightarrow 13 = 2C \Rightarrow C = \frac{13}{2}$ Therefore,$A+C = \frac{3}{2} + \frac{13}{2} = \frac{16}{2} = 8$

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C= \end{aligned}$

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