If frac{1}{x^4+x^2+1}=frac{A x+B}{x^2+x+1}+fr | vedclass

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(D) Given, $\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.Since $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$, we have:$1 = (Ax+B)(x^2-x+1) +

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(D) Given, $\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.Since $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$, we have:$1 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$.Expanding the right side:$1 = (Ax^3 - Ax^2 + Ax + Bx^2 - Bx + B) + (Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D)$.Grouping the terms by powers of $x$:$1 = (A+C)x^3 + (-A+B+C+D)x^2 + (A-B+C+D)x + (B+D)$.Comparing coefficients on both sides:$A+C = 0$ $(i)$$-A+B+C+D = 0$ (ii)$A-B+C+D = 0$ (iii)$B+D = 1$ (iv)Adding equations (ii) and (iii):$(-A+B+C+D) + (A-B+C+D) = 0 + 0$$2C + 2D = 0$$2(C+D) = 0$$C+D = 0$.

If $\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}$, then $C+D$ is equal to

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