Question

If $\frac{2x}{x^3-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$, then

• A. $A=-B=C$
• B. $A=B\ne C$
• C. $A\ne B=C$
• D. $A\ne B\ne C$

Answer 1

The correct option is A $A=-B=C$

Consider, $\frac{2x}{x^3-1}=\frac{2x}{(x-1)(x^2+x+1)}$
Resolving into partial fractions,
$\frac{2x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$\Rightarrow 2x=A(x^2+x+1)+(Bx+C)(x-1)$
$\Rightarrow 2x=(A+B)x^2+(A-B+C)x+(A-C)$
$\Rightarrow A+B=0,A-B+C=2,A-C=0$
$\Rightarrow A=-B=C$