Question

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

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

Answer 1

The correct option is D $1$

We have,

Cubic equation is $x^3-x-1=0$

Now, we can written as,

$x^3+0x^2-x-1=0$

Roots are $\alpha ,\beta and\gamma$

Then,

Sum of roots $=-\frac{coeff.of{x}^2}{coeff.of{x}^3}$

$\alpha +\beta +\gamma =\frac{-0}{1}$

$\alpha +\beta +\gamma =0......\left(1\right)$

Both product of roots$=\frac{coeff.ofx}{coeff.of{x}^3}$

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

$\alpha \beta +\beta \gamma +\gamma \alpha =-1......\left(2\right)$

Product of roots$=-\frac{\text{constant}\text{term}}{coeff.of{x}^3}$

$\alpha \beta \gamma =\frac{-1}{1}$

$\alpha \beta \gamma =-1......\left(3\right)$

From equation (1), (2) and (3) to and we get,

$\alpha =1,\beta =-1,\gamma =0$

Then,

According to question,

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

$\Rightarrow \frac{1+1}{1-1}+\frac{1-1}{1+1}+\frac{1+0}{1-0}$

$\Rightarrow 1$

Hence, this is he answer.