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If $\frac{1-x+6 x^2}{1-x^3}=\frac{A}{x}+\frac{B}{1+x}+\frac{C}{1+x}$, then $A$ is equal to
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Verified Answer
The correct answer is:
$1$
We have,
$$
\begin{aligned}
\frac{1-x+6 x^2}{x-x^3} & =\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x} \\
\Rightarrow \quad 1-x+6 x^2 & =A\left(1-x^2\right)+B x(1+x)+C x(1-x)
\end{aligned}
$$
Put $x=0$, then $A=1$
$$
\begin{aligned}
\frac{1-x+6 x^2}{x-x^3} & =\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x} \\
\Rightarrow \quad 1-x+6 x^2 & =A\left(1-x^2\right)+B x(1+x)+C x(1-x)
\end{aligned}
$$
Put $x=0$, then $A=1$
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