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Question

(1ω+ω2)(1ω2+ω4)(1ω4+ω8)(1ω8+ω16)=

A
4
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B
8
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C
12
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D
16
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Solution

The correct option is D 16

We have to find the value of (1ω+ω2)(1ω2+ω4)(1ω4+ω8)(1ω8+ω16)(1)

We know that since 1,ω,ω2are cube roots of unity

1+ω+ω2=0

And also

ω1=ω4==ω3n+1

ω2=ω5==ω3n+2

ω3=ω6==ω3n

1+ω2=ω

Now

(1ω+ω2)

=(1+ω2ω)

=2ω (1+ω2=ω)(2)

Also

(1ω2+ω4)=2ω2 (\because ω1=ω4==ω3n+1 &(1+ω=ω2)

For the third term

(1ω4+ω8)=(1ω1+ω2)

=2ω(3)

Similarly

Fourth term is

(1ω8+ω16)=2ω2(4)

Substituting values of (2),(3)&(4) in (1)

(2ω)(2ω2)(2ω)(2ω2)=16ω3ω3=16(ω3=1)

Hence option (D) is the correct answer


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