JEE Advance - Mathematics (1979 - No. 8)
If the cube roots of unity are $$1,\,\omega ,\,{\omega ^2},$$ then the roots of the equation $${\left( {x - 1} \right)^3} + 8 = 0$$ are
$$ - 1,1 + 2\omega ,\,1 + 2{\omega ^2}$$
$$ - 1,1 - 2\omega ,\,1 - 2{\omega ^2}$$
$$ - 1, - 1, - 1$$
None of these