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Question
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)3 + 8 = 0 are – 1, 1 – 2ω, 1 – 2ω2
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Solution
Given ω ≠ 1 is a active root of unity
(z – 1)3 + 8 = 0
(z – 1)3 = – 8
z – 1 = `(- 8)^(1/3) (1)^(1/3)`
= (– 2)(1, ω, ω2)
z – 1 = (– 2, – 2ω, – 2ω2)
= z – 1 = – 2
z = – 2 + 1 = – 1
z – 1 = – 2ω
z = 1 – 2ω
z – 1 = – 2ω2
z = 1 – 2ω2
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