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Question

Arrange the following values in ascending order:
I) (1+ω)(1+ω2)(1+ω4)(1+ω8)
II) (1ω+ω2)7+(1+ωω2)7
III) (1ω)(1ω2)(1ω4)(1ω8)
IV) (1ω+ω2)(1ω2+ω4)

A
I,IV,III,II
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B
I,II,III,IV
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C
I,III,IV,II
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D
IV,III,II,I
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Solution

The correct option is D I,IV,III,II
Using two formulae we can solve the problem , i.e.
1) 1+ω+ω2=0
2) ω3=1

I) (1+ω)(1+ω2)(1+ω4)(1+ω8)
=(ω2)(ω)(1+ω)(1+ω2)
=(ω2)(ω)(ω2)(ω)
=ω6=1

II) (1ω+ω2)7+(1+ωω2)7=128

III) (1ω)(1ω2)(1ω4)(1ω8)
=(1ω)(1ω2)(1ω)(1ω2)
=9

IV) (1ω+ω2)(1ω2+ω4)
=(ωω)(ω2ω2)
=4ω3=4
Hence, the correct sequence is I, IV, III, II.

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