lf a. b, c are the roots of ${\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}+26=0$ and w.r.t cube roots of unity then the value of $\frac{\mathrm{a}-1}{\mathrm{b}-1}+\frac{\mathrm{b}-1}{\mathrm{c}-1}+\frac{\mathrm{c}-1}{\mathrm{a}-1}= \_\_\_\_\_\_\_\_\_\_\_\_$

$\begin{array}{rcl}{\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}+26& =& 0\end{array}$

$\begin{array}{rcl}{\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}-1& =& -27\end{array}$

$\begin{array}{rcl}{(\mathrm{x}-1)}^3& =& -27\Rightarrow \mathrm{x}-1=-3, -3\mathrm{w}, -3{\mathrm{w}}^2\end{array}$

$\begin{array}{rcl}\mathrm{a}-1& =& -3, \mathrm{b}-1=-3\mathrm{w}, \mathrm{c}-1=-3{\mathrm{w}}^2\end{array}$

$\begin{array}{rcl} & & \\ & & \frac{\mathrm{a}-1}{\mathrm{b}-1}+\frac{\mathrm{b}-1}{\mathrm{c}-1}+\frac{\mathrm{c}-1}{\mathrm{a}-1}\\ & =& \frac{-3}{-3\mathrm{w}}+\frac{(-3\mathrm{w})}{-3{\mathrm{w}}^2}+\frac{(-3{\mathrm{w}}^2)}{-3}\\ & =& \frac{1}{\mathrm{w}}+\frac{1}{\mathrm{w}}+\frac{1}{\mathrm{w}}\\ & =& \frac{3}{\mathrm{w}}=3{\mathrm{w}}^2\end{array}$