Question

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

• A. $-7$
• B. $-6$
• C. $-5$
• D. $-4$

Answer 1

The correct option is A $-7$

$x^3-x-1=0$

$\Rightarrow$ $x(x^2-1)=1$

$\Rightarrow$ $x(x-1)(x+1)=1$

$\Rightarrow$ $-x(x+1)=\frac{1}{1-x}$

$\Rightarrow$ $\sum \left(\frac{1+\alpha }{1-\alpha }\right)=\sum -\alpha (\alpha +1{)}^2$

$\Rightarrow$ $\sum \left(\frac{1+\alpha }{1-\alpha }\right)=-\sum ({\alpha }^3+2{\alpha }^2+a)$ ----- ( 1 )

Now, we will calculate $\sum a^3$

Substituting $\alpha ,\beta ,\gamma$ in the equation and adding we get,

$\Rightarrow$ $\sum {\alpha }^3=\sum \alpha +3=0+3=3$ [ As sum of roots $\sum a=0$ ]

Now, we have to find $\sum {\alpha }^2$

$\Rightarrow$ $\sum {\alpha }^2=(\sum a{)}^2-2(\sum \alpha \beta )$

$\Rightarrow$ $\sum {\alpha }^2=0-2(-1)=2$

Now, we will substutute above value in equation ( 1 ) we get,

$\Rightarrow$ $=\sum \left(\frac{1+\alpha }{1-\alpha }\right)=-\sum ({\alpha }^3+2{\alpha }^2+a)=-(3+2(2)+0)=-7$