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

Question

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

$ - 1, - 1, - 1$

$ - 1,1 - 2\omega ,1 - 2{\omega ^2}$

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

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