Question

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

• A. $-7$
• B. $5$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

$\frac{1}{3}$

$\alpha ,\beta ,\gamma$ are roots of $x^3+x+1=0$
then $\alpha +1,\beta +1,\gamma +1$ are roots of ${(x-1)}^3+(x-1)+1=0$
$\Rightarrow x^3-1-3x^2+3x+x-1+1=0\Rightarrow x^3-3x^2+4x-1=0$
And $(\alpha +1)(\beta +1)(\gamma +1)=1$
Similarly $\alpha -1,\beta -1,\gamma -1$ are roots of ${(x+1)}^3+(x+1)+1=0$
$\Rightarrow x^3+1+3x^2+3x+x+1+1=0\Rightarrow x^3+3x^2+4x+3=0$
$\therefore (\alpha -1)(\beta -1)(\gamma -1)=-3$
Now $\prod \left(\frac{1+\alpha }{1-\alpha }\right)=-\left(\frac{1}{-3}\right)=\frac{1}{3}$