if 1,ω,ω² are three cube roots of unity, prove that (1+5ω²+ω⁴)

1.	if 1,ω,ω² are three cube roots of unity, prove that (1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)=64
Answer» Formulas to be used: 1) 1+ω+ω²=0 2) ω^3k=1,ω^3k+1=ω,ω^3k+2=ω² So,(1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²) (1+5ω²+ω)(1+5ω+ω²)(5+ω+ω²) So, 1+ω=-ω²,1+ω²=-ω, ω+ω²=-1 (5ω²-ω²)(5ω-ω)(5-1) (4ω²)(4ω)(4) 64ω³=64 Hence Proved (1 + 5ω² + ω⁴) (1 + 5ω + ω²) ( 5 + ω + ω²) = (1 + ω + ω²+ 4ω²) ( 1 + ω + ω² + 4ω) ( 4 +1 + ω + ω²) (∵ ω³= 1 ⇒ ω⁴ = ω) = 4ω².4ω.4 (∵ 1 + ω + ω² = 0) = 64ω³ = 64 (∵ ω³ = 1)

if 1,ω,ω² are three cube roots of unity, prove that (1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)=64

Answer»

Formulas to be used:

1) 1+ω+ω²=0

2) ω^3k=1,ω^3k+1=ω,ω^3k+2=ω²

So,(1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)

(1+5ω²+ω)(1+5ω+ω²)(5+ω+ω²)

So, 1+ω=-ω²,1+ω²=-ω, ω+ω²=-1

(5ω²-ω²)(5ω-ω)(5-1)

(4ω²)(4ω)(4)

64ω³=64

Hence Proved

(1 + 5ω² + ω⁴) (1 + 5ω + ω²) ( 5 + ω + ω²)

= (1 + ω + ω²+ 4ω²) ( 1 + ω + ω² + 4ω) ( 4 +1 + ω + ω²) (∵ ω³= 1 ⇒ ω⁴ = ω)

= 4ω².4ω.4 (∵ 1 + ω + ω² = 0)

= 64ω³ = 64 (∵ ω³ = 1)

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if 1,ω,ω² are three cube roots of unity, prove that (1+5ω²+ω⁴)(1+5ω+ω²)(5+ω+ω²)=64

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