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If a,b,c are integers not all equal and $ω$ is a cube root of unity $(ω ≠ 1)$ then the minimum value of $a + b ω + c ω 2$ is

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a+b+c

answer is C.

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Detailed Solution

Consider

a + b ω + c ω 2 2 = a + b ω + c ω 2 a + b ω + c ω 2 ¯

= a + b ω + c ω 2 a + b ω ¯ + c ω 2 ¯ = a + b ω + c ω 2 a + b ω 2 + c ω = a 2 + b 2 + c 2 − a b − b c − c a = 12 (a − b) 2 + (b − c) 2 + (c − a) 2

When a=b=1,c=2, it gives minimum value (Since a,b,c not all Equal)

∴

Minimum value of

a + b ω + c ω 2 = 12 (0 + 1 + 1) = 1 = 1

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