If frac{1}{x^4+x^2+1}=frac{Ax+B}{x^2+x+1}+fra | vedclass

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(B) Given,$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Equating the numerators:$1

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(B) Given,$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.Equating the numerators:$1 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$$1 = Ax^3 - Ax^2 + Ax + Bx^2 - Bx + B + Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D$$1 = (A+C)x^3 + (B-A+C+D)x^2 + (A-B+C+D)x + (B+D)$Comparing coefficients on both sides:$A+C = 0 \Rightarrow C = -A$$B-A+C+D = 0 \Rightarrow B+D = A-C = 2A$$A-B+C+D = 0$ $\Rightarrow A+C = B-D$ $\Rightarrow 0 = B-D$ $\Rightarrow B=D$$B+D = 1$ $\Rightarrow 2B = 1$ $\Rightarrow B = \frac{1}{2}, D = \frac{1}{2}$Since $B+D = 2A$,we have $1 = 2A \Rightarrow A = \frac{1}{2}$.Since $C = -A$,we have $C = -\frac{1}{2}$.Thus,$A+B+C+D = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 1$.Therefore,$\cos^{-1}(A+B+C+D) = \cos^{-1}(1) = 0$.

If $\frac{1}{x^4+x^2+1}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$,then $\cos^{-1}(A+B+C+D)=$

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