1
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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
\N
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B
1
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C
32
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D
12
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Solution

## The correct option is B 1Given that a, b, c are integers not all equal, ω is cube root of unity ≠ 1. Then a+bω+cω2=∣∣a+b(−1+i√32)+c(−1−i√32)∣∣=∣∣a+b(2a−b−c2)+i(b√3−c√32)∣∣=12√(2a−b−c)2+3(b−c)2=12√(4a2+b2+c2−4ab+2bc−4ac+3b2+3c2−6bc)=√a2+b2+c2−ab−bc−ca=√12[(a−b)2+(b−c)2+(c−a)2] R.H.S. will be minmum when a=b=c, but we cannot take a=b=c as per the question. Hence, the minimum value is obtained when any two are zero and third is a minimum magnitude integer, i.e., 1. Thus, b=c=0, a=1 gives us the minimum value of 1.

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