If $A + B + C = π$ and $\cos A + \cos B + \cos C = 0 = \sin A + \sin B + \sin C$ , then $\cos 3 A + \cos 3 B + \cos 3 C =$

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-3

answer is B.

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Detailed Solution

$\begin{array}{l} a = c i s A, b = c i s B, c = c i s C \\ a + b + c = c i s A + c i s B + c i s C \\ = (\cos A + \cos B + \cos C) + i (\sin A + \sin B + \sin C) \\ a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c \\ (cis A)^{3} + (cis B)^{3} + (cis C)^{3} = 3 cisAcisBcisC \\ c i s 3 A + c i s 3 B + c i s 3 C = 3 c i s (A + B + C) \\ (\cos 3 A + \cos 3 B + \cos 3 C) + i (\sin 3 A + \sin 3 B + \sin 3 C) \\ = 3 \cos (A + B + C) + i 3 \sin (A + B + C) \end{array}$

Equating real and imaginary part

$\begin{array}{l} \cos 3 A + \cos 3 B + \cos 3 C = 3 \cos (A + B + C) \\ \cos 3 A + \cos 3 B + \cos 3 C = 3 \cos π \\ \cos 3 A + \cos 3 B + \cos 3 C = - 3 \end{array}$