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Question

If x=3+i32 is a complex number, then the value of (x2+3x)2(x2+3x+1) is

A
98
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B
6
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C
18
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D
36
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Solution

The correct option is D 18
Given, x=3+i32
x=3±i32
x=1+i321=ω1
(x2+3x)2(x2+3x+1)
=[(ω1)2+3]2[(ω1)2+3(ω1)+1]
=(ω2+ω2)2(ω2+ω1)
[1+ω+ω2=0]
=(12)2(11)=18

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