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, $x^3-2x+1=0,(\alpha ,\beta ,\gamma )$ 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
$\Sigma \frac{1}{\alpha +\beta -\gamma }=\Sigma \frac{1}{-\gamma -\gamma }=-\frac{1}{2}\Sigma \frac{1}{\gamma }$
$=-\frac{1}{2}\left(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\right)$
$=-\frac{1}{2}\left(\frac{\alpha \beta +\beta \gamma +\alpha \gamma }{\alpha \beta \gamma }\right)$
$=-\frac{1}{2}\cdot \frac{\Sigma \alpha \beta }{\alpha \beta \gamma }=-\frac{1}{2}\cdot \frac{(-2)}{(-1)}=-1$