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Question

If ω is cube root of unity and x+y+z=a, x+ωy+ω2z=b, x+ω2y+ωz=b then which of the following is not correct?

A
x=a+b+c3
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B
y=a+bω2+ωc3
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C
x=a+bω+ω2c3
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D
None of these
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Solution

The correct option is D None of these
Given x+y+z=a...(i)
x+ωy+ω2z=b ...(ii)
x+ω2y+ωz=c ...(iii)
By adding (i), (ii) and (iii), we get
x=a+b+c3
Hence option A is correct.
Again (i)+(ii)×ω2+(iii)×ω, we get
3y=aω3+bω2+cω
y=a+bω2+cω3
Hence, option B is correct.
Similarly, (i)+(ii)×ω+(iii)×ω2 we get
z=a+bω+cω23
Hence, option C is correct

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