Question

Resolve $\frac{x}{1-x^3}$ into partial fraction.

Answer 1

$\frac{x}{1-x^3}=\frac{x}{(1-x)(1+x+x^2)}$
$\Rightarrow \frac{x}{1-x^3}=\frac{A}{1-x}+\frac{Bx+C}{1+x+x^2}$ .....(1)
$\Rightarrow x=A(1+x+x^2)+(Bx+C)(1-x)$ ....(2)
Putting $1-x=0\Rightarrow x=1$ in equation (2) we get
$\Rightarrow 1=A(1+1+1)+0$
$\Rightarrow 1=3A$
$\Rightarrow A=\frac{1}{3}$ ....(3)
Putting $x=0$ in equation (2) we get,
$\Rightarrow 0=A(1+0+0)+(B\times 0+C)(1-0)$
$\Rightarrow 0=A+C$
Putting the value of A from equation (3) we get
$\Rightarrow 0=\frac{1}{3}+C$
$C=-=\frac{1}{3}$ ........(4)
Comparing the coefficients of $x^2$ in both the sides of equation (2) we get.
$\Rightarrow O=A-B$
$\Rightarrow B=A$
Putting the value of $A$ from equation (3)
$B=\frac{1}{3}$ ....(5)
Putting the value of $A,B,C$ in equation (1) we get,
$\frac{x}{1-x^3}=\frac{1}{3(1-x)}+\frac{x-1}{3(1+x+x^2)}$.