Question

If $\frac{1}{x^3-2x^2-x+2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}$, then find the value of $A+B+C$.

Answer 1

$\frac{1}{x^3-2x^2-x+2}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{X}{(x-2)}$.
$\Rightarrow \frac{A(x+1)(x-2)+B(x-1)(x-2)+C(x-1)(x+1)}{(x-1)(x+1)(x-2)}$
Let $x-1=0$
$x=1$ then $1=A(1+1)(1-2)$
$1=A(-2)$
$A=-\frac{1}{2}$
Again $x+1=0$
$x=-1$
We get $1=B(-1-1)(-1-2)$
$1=B(-2)(-3)$
$1=6B$
$B=\frac{1}{6}$
again $x-2=0$
$x=2$
$1=C(2-1)(2+1)$
$1=C\times 1\times 3$
$1=3C$
$C=\frac{1}{3}$.
Find the value of $A+B+C$

$\Rightarrow \frac{-1}{2}+\frac{1}{6}+\frac{1}{3}=\frac{-3+1+2}{6}=\frac{0}{6}=0$.