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

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(B) We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Comparing this with $(x^2+ax+1)(x^2-ax+1) = x^4+(2-a^2)x^2+1$, we get $2-a^2=1$, which implies $a^2

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$2a$

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(B) We know that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$. Comparing this with $(x^2+ax+1)(x^2-ax+1) = x^4+(2-a^2)x^2+1$, we get $2-a^2=1$, which implies $a^2=1$. Thus, $a=1$.Given $\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+ax+1} + \frac{Cx+D}{x^2-ax+1}$.Multiplying both sides by $(x^2+ax+1)(x^2-ax+1)$, we get $1 = (Ax+B)(x^2-ax+1) + (Cx+D)(x^2+ax+1)$.Expanding the $RHS$: $1 = (A+C)x^3 + (-aA+B+aC+D)x^2 + (A-aB+C+aD)x + (B+D)$.Comparing coefficients:$1$) $A+C=0 \Rightarrow C=-A$$2$) $B+D=1$$3$) $-aA+B+aC+D = 0 \Rightarrow -aA+B-aA+D = 0 \Rightarrow -2aA + (B+D) = 0 \Rightarrow -2aA+1=0 \Rightarrow A = \frac{1}{2a}$. Thus $C = -\frac{1}{2a}$.$4$) $A-aB+C+aD = 0 \Rightarrow (A+C) - a(B-D) = 0 \Rightarrow 0 - a(B-D) = 0 \Rightarrow B=D$.Since $B+D=1$ and $B=D$, we have $B=D=\frac{1}{2}$.Now, $A+B-C+D = \frac{1}{2a} + \frac{1}{2} - (-\frac{1}{2a}) + \frac{1}{2} = \frac{1}{a} + 1$.Since $a=1$, $A+B-C+D = 1+1 = 2 = 2a$.

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

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