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If $F_1$ and $F_2$ are irreducible factors of $x^4+x^2+1$ with real coefficients and $\frac{x^3-2 x^2+3 x-4}{x^4+x^2+1}=\frac{A x+B}{F_1}+\frac{C x+D}{F_2}$, then $A+B+C+D=$
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Verified Answer
The correct answer is:
-3
We have,
$$
\begin{aligned}
x^4+x^2+1 & =x^4+x^2+x^2-x^2+1 \\
& =x^4+2 x^2+1-x^2=\left(x^2+1\right)^2-(x)^2 \\
& =\left(x^2+x+1\right)\left(x^2-x+1\right)
\end{aligned}
$$
Now,
$$
\begin{aligned}
& \begin{array}{l}
\frac{x^3-2 x^2+3 x-4}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1} \\
\Rightarrow x^3-2 x^2+3 x-4=(A x+B)
\end{array} \\
& \qquad \quad\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right) \\
& \Rightarrow x^3-2 x^2+3 x-4=(A+C) x^3 \\
& +x^2(B-A+C+D)+x(A-B+C+D)+(B+D)
\end{aligned}
$$
Comparing of all like terms
$$
\begin{aligned}
A+C & =1 \\
B-A+C+D & =-2 \\
A-B+C+D & =3 \\
B+D & =-4
\end{aligned}
$$
Adding Eqs. (i) and (ii), we get
$$
A+B+C+D=1+(-4)=-3
$$
$$
\begin{aligned}
x^4+x^2+1 & =x^4+x^2+x^2-x^2+1 \\
& =x^4+2 x^2+1-x^2=\left(x^2+1\right)^2-(x)^2 \\
& =\left(x^2+x+1\right)\left(x^2-x+1\right)
\end{aligned}
$$
Now,
$$
\begin{aligned}
& \begin{array}{l}
\frac{x^3-2 x^2+3 x-4}{x^4+x^2+1}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{x^2-x+1} \\
\Rightarrow x^3-2 x^2+3 x-4=(A x+B)
\end{array} \\
& \qquad \quad\left(x^2-x+1\right)+(C x+D)\left(x^2+x+1\right) \\
& \Rightarrow x^3-2 x^2+3 x-4=(A+C) x^3 \\
& +x^2(B-A+C+D)+x(A-B+C+D)+(B+D)
\end{aligned}
$$
Comparing of all like terms
$$
\begin{aligned}
A+C & =1 \\
B-A+C+D & =-2 \\
A-B+C+D & =3 \\
B+D & =-4
\end{aligned}
$$
Adding Eqs. (i) and (ii), we get
$$
A+B+C+D=1+(-4)=-3
$$
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