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Question

If 1,ω,ω2 are the cube roots of unity, prove that:
(a+b)(aω+bω2)(aω2+bω)=a3+b3

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Solution

Given:- 1,ω,ω2 are the cube roots of unity
To prove:- (a+b)(aω+bω2)(aω2+bω)=a3+b3
Proof:-
Since 1,ω and ω2 are the cube roots of unity.
Therefore,
1+ω+ω2=0
ω3=1
ω3n+1=ω
Now,
(a+b)(aω+bω2)(aω2+bω)
=(a2ω+b2ω2+ab(ω+ω2))(aω2+bω)
=(a2ω+b2ω2ab)(aω2+bω)
=a3ω3+a2bω2+b2aω4+b3ω3a2bω2ab2ω
=a3a2bω2+ab2ω+b3a2bω2ab2ω
=a3+b3
Hence proved.

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