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Question

If 1,ω,ω2 are the cube roots of unity, then prove that (2ω)(2ω2)(2ω10)(2ω11)=49

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Solution

Given 1,ω,ω2 are the cube roots of unity
1+ω+ω2=0,ω3=1
LHS=(2ω)(2ω2)(2ω10)(2ω11)=(2ω)(2ω2)(2(ω3)3ω)(2(ω3)3ω2)=((2ω)(2ω)2)2
=(222(ω+ω2)+ω3)2=(4+2+1)2=72=49=RHS
Hence proved

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