MathematicsEasy53×since 2002Q3231If the cube roots of unity are 1, ω , ω2\omega \,,\,{\omega ^2}ω,ω2 then the roots of the equation (x−1)3{(x - 1)^3}(x−1)3 + 8 = 0, are :A−1,−1+2 ω,−1−2 ω2- 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}−1,−1+2ω,−1−2ω2B−1,−1,−1- 1, - 1, - 1−1,−1,−1C−1,1−2ω,1−2ω2- 1,1 - 2\omega ,1 - 2{\omega ^2}−1,1−2ω,1−2ω2D−1,1+2ω,1+2ω2- 1,1 + 2\omega ,1 + 2{\omega ^2}−1,1+2ω,1+2ω2Check answerSkip