If the cube roots of unity are $1, \omega, \omega^2$,then the roots of the equation $(x - 2)^3 + 27 = 0$ are
- A$ - 1, - 1, - 1$
- B$ - 1, - \omega, - \omega^2$
- C$ - 1, 2 + 3\omega, 2 + 3\omega^2$
- D$ - 1, 2 - 3\omega, 2 - 3\omega^2$
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DifficultKVPY 2010
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