Question

Resolve into partial fraction $\frac{x^3-3x-2}{(x^2+x+1)(x+1{)}^2}$

• A. $\frac{3x-1}{x^2+x+1}+\frac{2}{(x+1{)}^2}-\frac{3}{(x+1)}$
• B. $\frac{3x-1}{x^2+x+1}+\frac{2}{(x+1{)}^2}+\frac{3}{(x+1)}$
• C. $\frac{3x}{x^2+x+1}+\frac{2}{(x+1{)}^2}-\frac{3}{(x+1)}$
• D. $\frac{x-1}{x^2+x+1}+\frac{2}{(x+1{)}^2}-\frac{3}{(x+1)}$

Answer 1

The correct option is A $\frac{3x-1}{x^2+x+1}+\frac{2}{(x+1{)}^2}-\frac{3}{(x+1)}$

Let $\frac{x^3-3x-2}{(x^2+x+1){(x+1)}^2}=\frac{Ax+B}{x^2+x+1}+\frac{C}{x+1}+\frac{D}{{(x+1)}^2}$
$\Rightarrow x^3-3x-2=(Ax+B){(x+1)}^2+C(x^2+x+1)(x+1)+D(x^2+x+1)\Rightarrow x^3-3x-2=A(x^3+2x^2+2x)+B(x^2+2x+1)+C(x^3+2x^2+2x+1)+D(x^2+x+1)$
On comapring coefficients we get
$A+C=1,2A+B+2C+D=0,2A+2B+2C+D=-3,B+C+D=-2\Rightarrow A=3,B=-1,C=-3,D=2$
Hence
$\frac{x^3-3x-2}{(x^2+x+1){(x+1)}^2}=\frac{3x-1}{x^2+x+1}+\frac{-3}{x+1}+\frac{2}{{(x+1)}^2}$
Hence, option 'A' is correct.