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Question

If x=a+b,y=aω+bω2,z=aω2+bω, then xyz equals to where, ω is the cube root of unity.

A
a+b
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B
a2+b2
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C
a3+b3
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D
a4+b4
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Solution

The correct option is C a3+b3
Given : x=a+b,y=aω+bω2,z=aω2+bω
Consider, xyz
=(a+b)(aω+bω2)(aω2+bω)
=(a2ω+abω2+abω+b2ω2)(aω2+bω)
=a3ω3+a2bω2+a2bω4+ab2ω3+a2bω3+ab2ω2+ab2ω4+b3ω3
=a3+b3+a2b(ω2+ω4+ω3)+ab2(ω3+ω2+ω4) ....... [ω3=1]
=a3+b3+a2b(ω2+ω+1)+ab2(ω2+ω+1)
=a3+b3 ..... [(ω2+ω+1)=0,ω3=1]

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