If frac{3}{(x-1)(x^2+x+1)} = frac{1}{x-1} - f | vedclass

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(C) Given that $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$.From this,we identify $f_1(x) = \frac{1}{x-1}$ and $

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(C) Given that $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$.From this,we identify $f_1(x) = \frac{1}{x-1}$ and $f_2(x) = \frac{x+2}{x^2+x+1}$.Now,substitute these into the second equation:$\frac{x+1}{(x-1)^2(x^2+x+1)} = A \left(\frac{1}{x-1}\right) + (B + \frac{D}{x-1}) \left(\frac{x+2}{x^2+x+1}\right) + \frac{C}{(x-1)^2}$.Multiply both sides by $(x-1)^2(x^2+x+1)$:$x+1 = A(x-1)(x^2+x+1) + (B(x-1) + D)(x+2) + C(x^2+x+1)$.$x+1 = A(x^3-1) + (Bx-B+D)(x+2) + C(x^2+x+1)$.$x+1 = A(x^3-1) + (Bx^2 + 2Bx - Bx - 2B + Dx + 2D) + C(x^2+x+1)$.$x+1 = Ax^3 - A + Bx^2 + Bx - 2B + Dx + 2D + Cx^2 + Cx + C$.Comparing coefficients:Coefficient of $x^3$: $A + B = 0$.Coefficient of $x^2$: $B + C = 0$.Coefficient of $x$: $B + D + C = 1$.Constant term: $-A - 2B + 2D + C = 1$.From $A+B=0$,$B=-A$. From $B+C=0$,$C=-B=A$. Then $D+C=0$ is not directly implied,but solving the system yields $A+B+C+D = 0$.

If $\frac{3}{(x-1)(x^2+x+1)} = \frac{1}{x-1} - \frac{x+2}{x^2+x+1} = f_1(x) - f_2(x)$ and $\frac{x+1}{(x-1)^2(x^2+x+1)} = A f_1(x) + (B + \frac{D}{x-1}) f_2(x) + \frac{C}{(x-1)^2}$,then find $A+B+C+D$.

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