Question

If $\alpha$, $\beta$ and $\gamma$ are the roots of the equation $x^3-x-1=0$, then $\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }$ has the value

• A. $-5$
• B. $-1$
• C. $-7$
• D. $1$

Answer 1

The correct option is A $-5$

As $\alpha ,\beta ,\gamma aretherootsof{x}^3-x^2-1=0$

So, $\alpha +\beta +\gamma =1,\alpha \beta \gamma =1and\alpha \beta +\beta \gamma +\gamma \alpha =0$.......(i)

$[As,\alpha +\beta +\gamma =\frac{-b}{a},\alpha \beta \gamma =\frac{-d}{a}and\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}]$

Now, $\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }$

$=\frac{(1+\alpha )(1-\beta )(1-\gamma )+(1-\alpha )(1+\beta )(1-\gamma )+(1-\alpha )(1-\beta )(1+\gamma )}{(1-\alpha )(1-\beta )(1-\gamma )}$

$=\frac{3-(\alpha +\beta +\gamma )-(\alpha \beta +\beta \gamma +\gamma \alpha )+3\left(\alpha \beta \gamma \right)}{1-(\alpha +\beta +\gamma )-(\alpha \beta +\beta \gamma +\gamma \alpha )+\left(\alpha \beta \gamma \right)}$

$=\frac{3-\left(1\right)-\left(0\right)+3\left(1\right)}{1-\left(1\right)+\left(0\right)-\left(1\right)}$

$=\frac{5}{-1}$

$=-5$