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If the equivalent partial fraction of $\frac{x^3}{(2 x-1)(x+2)(x-3)}$
is of the form $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$ then the value of $A+B+C=$
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is of the form $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$ then the value of $A+B+C=$
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Verified Answer
The correct answer is:
$4 / 25$
$\frac{x^3}{(2 x-1)(x+2)(x-3)}$
$=A+\frac{B}{(2 x-1)}+\frac{C}{(x+2)}+\frac{D}{(x-3)}$
$\begin{aligned} & x^3=A(2 x-1)(x+2)(x-3)+\mathrm{B}(x+2)(x-3)+\mathrm{C} \\ & (2 x-1)(x-3)+\mathrm{D}(2 x-1)(x+2) \ldots(\mathrm{i})\end{aligned}$
Put $x=1 / 2$
$\frac{1}{8}=B\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-3\right) \Rightarrow B=\frac{-1}{50}$
Put $x=-2$
$8=C(-4-1)(-2-3) \Rightarrow C=8 / 25$
Put $x=3$
$9=D(6-1)(3+2) \Rightarrow D=9 / 25$
Comparing coefficient of $x^3$ both sides in (i)
$1=2 \mathrm{~A} \Rightarrow \mathrm{A}=1 / 2$
$A+B+C=\frac{1}{2}-\frac{1}{50}+\frac{8}{25}$
$=\frac{25-1-16}{50}=\frac{8}{50}=\frac{4}{25}$
$=A+\frac{B}{(2 x-1)}+\frac{C}{(x+2)}+\frac{D}{(x-3)}$
$\begin{aligned} & x^3=A(2 x-1)(x+2)(x-3)+\mathrm{B}(x+2)(x-3)+\mathrm{C} \\ & (2 x-1)(x-3)+\mathrm{D}(2 x-1)(x+2) \ldots(\mathrm{i})\end{aligned}$
Put $x=1 / 2$
$\frac{1}{8}=B\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-3\right) \Rightarrow B=\frac{-1}{50}$
Put $x=-2$
$8=C(-4-1)(-2-3) \Rightarrow C=8 / 25$
Put $x=3$
$9=D(6-1)(3+2) \Rightarrow D=9 / 25$
Comparing coefficient of $x^3$ both sides in (i)
$1=2 \mathrm{~A} \Rightarrow \mathrm{A}=1 / 2$
$A+B+C=\frac{1}{2}-\frac{1}{50}+\frac{8}{25}$
$=\frac{25-1-16}{50}=\frac{8}{50}=\frac{4}{25}$
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