$$\begin{gathered} If a,b,c are distinct roots of x^3-3x^2+3x+26=0 and w.r.t cube roots of unity then the value of \\ \left|\frac{a-1}{b-1}+\frac{b-1}{c-1}+\frac{c-1}{a-1}\right|= \end{gathered}$$

$$\begin{gathered} Given equation can be written as {(x-1)}^3={(-3)}^3 \Rightarrow x=1-3,1-3\omega ,1-3{\omega }^2 where {\omega }^3=1 \\ let a=-2,b=1-3\omega ,c=1-3{\omega }^2 \\ \Rightarrow a-1=-3,b-1=-3\omega ,c-1=-3{\omega }^2 \\ Now \left|\frac{a-1}{b-1}+\frac{b-1}{c-1}+\frac{c-1}{a-1}\right|=\left|\frac{-3}{-3\omega }+\frac{-3\omega }{-3{\omega }^2}+\frac{-3{\omega }^2}{-3}\right| \\ =\left|\frac{1}{\omega }+\frac{1}{{\omega }^{ }}+\frac{{\omega }^2}{1}\right| \\ =\left|\frac{3}{\omega }\right| \\ =\left|3{\omega }^2\right| \\ =3\left|\frac{-1-i\sqrt{3}}{2}\right| \\ =3\left(1\right)=3 \end{gathered}$$