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Question: If the cube roots of unity be \(1,\omega,\omega^{2}\), then the roots of the equation \((x - 1)^{3} ...

If the cube roots of unity be 1,ω,ω21,\omega,\omega^{2}, then the roots of the equation (x−1)3+8=0(x - 1)^{3} + 8 = 0 are

A

−1,1+2ω,1+2ω2- 1,1 + 2\omega,1 + 2\omega^{2}

B

−1,1−2ω,1−2ω2- 1,1 - 2\omega,1 - 2\omega^{2}

C

−1,1,−1- 1,1, - 1

D

None of these

Answer

−1,1−2ω,1−2ω2- 1,1 - 2\omega,1 - 2\omega^{2}

Explanation

Solution

Sol. (x−1)3=−8⇒x−1=(−8)1/3(x - 1)^{3} = - 8 \Rightarrow x - 1 = ( - 8)^{1/3} ⇒ x−1=−2,−2ω,−2ω2x - 1 = - 2, - 2\omega, - 2\omega^{2}

⇒ x=−1,1−2ω,1−2ω2x = - 1,1 - 2\omega,1 - 2\omega^{2}