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If the equivalent partial fraction of $\frac{x^3}{(2 x-1)(x+2)(x-3)}$ is given by. $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$, then the value of $C$ is
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Verified Answer
The correct answer is:
$-\frac{8}{25}$
Here,
$\begin{array}{l}
\frac{x^3}{(2 x-1)(x+2)(x-3)}=A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3} \\
\Rightarrow x^3=A(2 x-1)(x+2)(x-3)+B(x+2)(x-3) . \\
\quad+C(2 x-1)(x-3)+D(2 x-1)(x+2)
\end{array}$
$\begin{aligned} & \text {On putting } x=-2 \text {, } \\ & -8=C(-5)(-5) \\ & C=\frac{-8}{25} \\ & \end{aligned}$
$\begin{array}{l}
\frac{x^3}{(2 x-1)(x+2)(x-3)}=A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3} \\
\Rightarrow x^3=A(2 x-1)(x+2)(x-3)+B(x+2)(x-3) . \\
\quad+C(2 x-1)(x-3)+D(2 x-1)(x+2)
\end{array}$
$\begin{aligned} & \text {On putting } x=-2 \text {, } \\ & -8=C(-5)(-5) \\ & C=\frac{-8}{25} \\ & \end{aligned}$
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