Question

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

• A. $A = B = C$
• B. $A = B \neq C$
• C. $A \neq B = C$
• D. $A \neq B \neq C$

Answer 1

Answer: (D)

$A \neq B \neq C$

Answer 2

$2x = A(x^{2} + x + 1) + (Bx + C)(x - 1)$

For $x = 1$, $2 = 3A$ $\Rightarrow$ $A = \frac{2}{3}$

For $x = \omega,2\omega = A(1 + \omega + \omega^{2}) + B\omega^{2} + (C - B)\omega - C$

$\Rightarrow$ $2\omega = A.0 + B\omega^{2} + (C - B)\omega - C$

$\omega = \frac{- 1 + \sqrt{3}i}{2},\omega^{2} = \frac{- 1 - \sqrt{3}i}{2}$

$\therefore - 1 + \sqrt{3}i = B\left( - \frac{1}{2} - \frac{\sqrt{3}}{2}i \right) + (C - B)\left( - \frac{1}{2} + \frac{\sqrt{3}}{2}i \right) - C$

$\Rightarrow - 1 + \sqrt{3}i = \left( - \frac{B}{2} - \frac{C}{2} + \frac{B}{2} - C \right) + \frac{i\sqrt{3}}{2}(C - 2B)$

$\Rightarrow - 1 = - \frac{3}{2}C,\sqrt{3} = \frac{\sqrt{3}}{2}(C - 2B)$

$C = \frac{2}{3},B = \frac{C - 2}{2} = - \frac{2}{3}$

$\therefore A = C \neq B$ $\Rightarrow$ $A \neq B \neq C$.