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Question

# If a, b, c are integers, not all simultaneously equal and ω is a cube root of unity (ω≠1), then the minimum value of ∣∣a+bω+cω2∣∣ is

A
1
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B
32
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C
12
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Solution

## The correct option is A 1Given that a,b,c are integers not all equal, ω is cube root of unity ≠ 1So, 1+ω+ω2=0⇒ω2=−(1+ω)Consider ∣∣a+bω+cω2∣∣Substitute ω2=−(1+ω), we get∣∣a+bω+cω2∣∣=|a+bω+c[−(1+ω)]| =a+bω−c−cω=a−c+(b−c)ωSince ω=1±i√32 as ω≠1⇒(∣∣a+bω+cω2∣∣=∣∣∣a−c+(b−c)[1±i√32]∣∣∣=∣∣∣[(a−c)+b−c2]±i(b−c)√32∣∣∣=√[(a−c)+b−c2]2+[(b−c)√32]2Since, a,b,c are integers, not all simultaneously equal.Let b=c=k and a=k+1 then∣∣a+bω+cω2∣∣=√[0+1]2+02=12+0=1Hence, the minimum value of given expression is 1

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