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

$\begin{array}{l}(\mathrm{x}-1{)}^3+8=0\Rightarrow (\mathrm{x}-1{)}^3={\left(-2\right)}^3\\ \Rightarrow \mathrm{x}-1=-2;-2\mathrm{\omega };-2{\mathrm{\omega }}^2\end{array}$

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

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

$\therefore \text{ required expression }=3{\mathrm{\omega }}^2\text{. }$