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If ω is a complex cube root of unity, show that (2 + ω + ω2)3 - (1 - 3ω + ω2)3 = 65 - Mathematics and Statistics

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Question

If ω is a complex cube root of unity, show that (2 + ω + ω2)3 - (1 - 3ω + ω2)3 = 65

Sum
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Solution

ω is a complex cube root of unity.

∴ ω3 = 1 and 1 + ω + ω2 = 0

Also, 1 + ω2 = - ω, 1 + ω = - ω2 and ω + ω2 = - 1

L.H.S. = (2 + ω + ω2)3 - (1 - 3ω + ω2)3 

= [(2 + (ω + ω2)]3 - [(- 3ω + (1 + ω2)]3 

= (2 - 1)3 - (- 3ω - ω)3

= 13 - (- 4ω)3

= 1 + 64ω3

= 1 + 64(1) = 65

= R.H.S.

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Chapter 3: Complex Numbers - EXERCISE 3.3 [Page 42]

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