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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Aicharya · Accepted Answer

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ight)+B x(1+x)+C x(1-x) \end{aligned} $ Put $x=0$, then $A=1$

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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