Question

If $\alpha , \beta ,\gamma$ are the roots of $x^3-2x+1=0$, then the value of $(\sum \frac {1} {\alpha +\beta -\gamma}$) is

• A. $\frac {-1}{2}$
• B. $-1$
• C. $0$
• D. $\frac {1}{2}$

Answer 1

Answer: (B)

Given cubic equation is, $x^{3}-2 x+1=0,(\alpha, \beta, \gamma)$ are roots of this equation.
Then, sum of roots $\Sigma \alpha=0$
$\Rightarrow \alpha+\beta+\gamma=0$
$\Sigma \alpha \beta=-2, \alpha \beta \gamma=-1$
Now, we have
$\Sigma \frac{1}{\alpha+\beta-\gamma} =\Sigma \frac{1}{-\gamma-\gamma}=-\frac{1}{2} \Sigma \frac{1}{\gamma}$
$=-\frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right)$
$=-\frac{1}{2}\left(\frac{\alpha \beta+\beta \gamma+\alpha \gamma}{\alpha \beta \gamma}\right)$
$=-\frac{1}{2} \cdot \frac{\Sigma \alpha \beta}{\alpha \beta \gamma}=-\frac{1}{2} \cdot \frac{(-2)}{(-1)} =-1$