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If ω is a complex cube root of unity, show that (3 + 3ω + 5ω2)6 − (2 + 6ω + 2ω2)3 = 0

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प्रश्न

If ω is a complex cube root of unity, show that (3 + 3ω + 5ω2)6 − (2 + 6ω + 2ω2)3 = 0

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उत्तर

ω is a complex cube root of unity.

∴ ω3 = 1 and 1 + ω + ω2 = 0.

∴ ω + ω2 = – 1, 1 + ω = – ω2 and 1 + ω2 = – ω.

(3 + 3ω + 5ω2)6 − (2 + 6ω + 2ω2)3 

= [3(1 + ω) + 5ω2]6 –  [2(1 + ω2) + 6ω]3

= [3(– ω2) + 5ω2]6 - [2(– ω) + 6ω)3

= (2ω2)6 –  (4ω)3

= 64ω12 –  64ω3

= 64(ω3)4 – 64ω3

= 64(1)4  – 64(1)

= 0

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अध्याय 1: Complex Numbers - Exercise 1.4 [पृष्ठ २०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 1 Complex Numbers
Exercise 1.4 | Q 2. (v) | पृष्ठ २०

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