InterviewSolution
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InterviewSolution
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If ω ≠ 1 is a cube root of unity, show that (i) (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128 (ii) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)……(1 + ω2n) = 1 |
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Answer» (i) ω is a cube root of unity ω3 = 1; 1 + ω + ω2 = 0 (1 – ω + ω2)6 + (1 + ω – ω2)6 = (-ω – ω)6 + (-ω2– ω 2)6 = (-2ω)6 + (-2ω2)6 = (-2)6 (ω6 + ω12) = (64)(1 + 1) = 128 (ii) (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) …… (1 + ω2n) = (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) ……. 2n factors = (-ω2)(-ω)(-ω2)(-ω) …… 2n factors = ω3 . ω3 = 1 |
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