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Question

If a, b, c are integers, not all simultaneously equal and ω is a cube root of unity (ω1), 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 1
Given that a, b, c are integers not all equal, ω is cube root of unity
1. Then
a+bω+cω2=a+b(1+i32)+c(1i32)=a+b(2abc2)+i(b3c32)=12(2abc)2+3(bc)2=12(4a2+b2+c24ab+2bc4ac+3b2+3c26bc)=a2+b2+c2abbcca=12[(ab)2+(bc)2+(ca)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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