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

${\left(x-1\right)}^3=-8$

${\left(x-1\right)}^3-{\left(-2\right)}^3=0$        $\left(\because a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right)$

$\left(x+1\right)\left(x^2-4x+7\right)=0$

$\alpha =-1,\beta =2+i\sqrt{3},\gamma =2-i\sqrt{3}$

$\alpha -1=-2,\beta -1=1+i\sqrt{3},\gamma -1=1-i\sqrt{3}$

$\alpha -1=-2,\beta -1=-2\omega ,\gamma -1=-2{\omega }^2$

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

$\Rightarrow \frac{2}{\omega }+{\omega }^2=\frac{3}{\omega }\left(\because {\omega }^3=1\right)$