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If $\frac{x^2-x+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{\mathrm{A} x+\mathrm{B}}{x^2+1}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+x+1}$ then $\mathrm{A}+2 \mathrm{~B}+\mathrm{C}+2 \mathrm{D}=$
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The correct answer is:
2
We are given that
$\begin{aligned}
& \frac{x^2-x+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{A x+B}{x^2+1}+\frac{(x+D)}{x^2+x+1} \\
& \text { Now } x^2-x+1=(A x+B)\left(x^2+x+1\right)+(c x+D)\left(x^2+1\right) \\
& \Rightarrow x^2-x+1=(A+C) x^3+(A+B+D) x^2+(A+B+C) x+B+D \\
& A+C=0 \\
& A+B+D=1 \\
& A+B+C=-1 \\
& B+D=1 \\
& \Rightarrow A=0, \quad C=0, \quad B=-1, \quad D=2 \\
& \therefore A+2 B+C+2 D=0-2+0+4=2
\end{aligned}$
$\begin{aligned}
& \frac{x^2-x+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{A x+B}{x^2+1}+\frac{(x+D)}{x^2+x+1} \\
& \text { Now } x^2-x+1=(A x+B)\left(x^2+x+1\right)+(c x+D)\left(x^2+1\right) \\
& \Rightarrow x^2-x+1=(A+C) x^3+(A+B+D) x^2+(A+B+C) x+B+D \\
& A+C=0 \\
& A+B+D=1 \\
& A+B+C=-1 \\
& B+D=1 \\
& \Rightarrow A=0, \quad C=0, \quad B=-1, \quad D=2 \\
& \therefore A+2 B+C+2 D=0-2+0+4=2
\end{aligned}$
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