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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 $\cos ^{-1}(A+B+C+D)=$
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Given, $\frac{1}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1}$
$\Rightarrow \frac{1}{x^4+x^2+1}$
$=\frac{(A x+B)\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right)}{x^4+x^2+1}$
$\Rightarrow \mathrm{l}=(A+C) x^3+(B-A+C+D) x^2$ $+(A-B+C+D) x+(B+D)$
On comparing the terms, we get
$A+C=0, A=B+C+D, A+C+D=B$
and $B+D=1$
$\therefore \quad A+B+C+D=1$
Therefore, $\cos ^{-1}(A+B+C+D)=\cos ^{-1}(1)=0$
$\Rightarrow \frac{1}{x^4+x^2+1}$
$=\frac{(A x+B)\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right)}{x^4+x^2+1}$
$\Rightarrow \mathrm{l}=(A+C) x^3+(B-A+C+D) x^2$ $+(A-B+C+D) x+(B+D)$
On comparing the terms, we get
$A+C=0, A=B+C+D, A+C+D=B$
and $B+D=1$
$\therefore \quad A+B+C+D=1$
Therefore, $\cos ^{-1}(A+B+C+D)=\cos ^{-1}(1)=0$
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