If the cube roots of unity are $$1, \omega, \omega^2$$ then the roots of the equation $$(x - 1)^3 + 8 = 0$$ are

Let $$y = x - 1$$. Then the given equation $$(x-1)^3 + 8 = 0$$ becomes $$y^{3} = -8$$.

Write $$-8$$ as $$2^{3}\,(-1)$$. Hence $$y^{3} = 2^{3}\,(-1)$$.

We need the three cube-roots of $$-1$$. If $$1,\,\omega,\,\omega^{2}$$ are the cube-roots of unity, then $$\omega^{3} = 1,\qquad 1+\omega+\omega^{2}=0,\qquad \omega \neq 1.$$ Expressing $$-1$$ in terms of these roots:

$$-1, \; -\omega, \; -\omega^{2}$$ are the three distinct numbers whose cubes give $$-1$$, because $$(-1)^{3}=-1,\quad (-\omega)^{3} = -\omega^{3} = -1,\quad (-\omega^{2})^{3} = -\omega^{6} = -1.$$

Therefore the three cube-roots of $$-8$$ are obtained by multiplying each cube-root of $$-1$$ by $$2$$:

$$y = 2(-1),\; 2(-\omega),\; 2(-\omega^{2}) \; \Longrightarrow \; y = -2,\; -2\omega,\; -2\omega^{2}.$$

Finally, convert back to $$x$$ using $$x = y + 1$$:

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

Hence the roots of the equation are $$-1,\; 1 - 2\omega,\; 1 - 2\omega^{2}$$.

Option C which is: $$-1,\; 1 - 2\omega,\; 1 - 2\omega^{2}$$