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