Question

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

• A. $-5$
• B. $-6$
• C. $-7$
• D. $-2$

Answer 1

The correct option is A $-5$

Let $y=\frac{1+\alpha }{1-\alpha }\Rightarrow \alpha =\frac{y-1}{y+1}$
As $\alpha ,\beta ,\gamma$ are roots of $x^3-x^2-1=0$
Replacing $x\to \frac{y-1}{y+1}$ in above equation, we get
${\left(\frac{y-1}{y+1}\right)}^3-{\left(\frac{y-1}{y+1}\right)}^2-1=0\Rightarrow {(y-1)}^3-{(y-1)}^2(y+1)-{(y+1)}^3=0\Rightarrow y^3-1-3y^2+3y-y^3+y+2y^2-y^2-1+2y-y^3-1-3y^2-3y=0\Rightarrow -y^3-5y^2+3y-3=0\Rightarrow y^3+5y^2-3y+3=0$
And roots are $\frac{1+\alpha }{1-\alpha },\frac{1+\beta }{1-\beta },\frac{1+\gamma }{1-\gamma }$
Hence, $\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\gamma }{1-\gamma }=-5$
Hence, option 'A' is correct.