Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3-3x^2+3x+7=0$ and $\omega$ is a cube root of unity, then $\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=$

• A. ${\omega }^2$
• B. $2{\omega }^2$
• C. $3{\omega }^2$
• D. $-3{\omega }^2$

Answer 1

The correct option is C $3{\omega }^2$

$x^3-3x^2+3x+7=0$
$\Rightarrow x^3-3x^2+3x-1+8=0$
$\Rightarrow (x-1{)}^3=-8$
$\Rightarrow {\left(\frac{x-1}{-2}\right)}^3=1$
$\Rightarrow \frac{x-1}{-2}=(1{)}^{1/3}$
$\Rightarrow \frac{x-1}{-2}=1,\omega ,{\omega }^2$
$\therefore x=-1,1-2\omega ,1-2{\omega }^2$
$\therefore \frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}$
$=\frac{-2}{-2w}+\frac{-2\omega }{-2{\omega }^2}+\frac{-2{\omega }^2}{-2}$
$=\frac{1}{\omega }+\frac{1}{\omega }+\frac{1}{\omega }=\frac{3}{\omega }=3{\omega }^2$