If the cube roots of unity are $1,\omega ,{\omega }^2$then the sum of roots of ${(x-1)}^3+27=0\Rightarrow {(x-1)}^3=-27$ is

${(x-1)}^3+27=0\Rightarrow {(x-1)}^3=-27$

$\Rightarrow (x-1)={(-27)}^{\frac{1}{3}}={(-27)}^{\frac{1}{3}}.{\left(1\right)}^{\frac{1}{3}}$

$\Rightarrow x-1=-3(1,\omega ,{\omega }^2)$

$\therefore \Rightarrow x-1=-3\omega ,-3,-3{\omega }^2$

$\Rightarrow x=1-3,1-3\omega ,1-3{\omega }^2$

$=-2, 1-3\omega +1-3{\omega }^2$

$\therefore$ sum of roots $=-2+2-3(\omega +{\omega }^2)$

$=0-3\left[-1\right]$          $\left(\because \hspace{0.17em} 1 + \omega + \omega 2 = 0, \omega + \omega 2 = - 1\right)$

$=0+3=3$