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

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$3 \cos (A + B + C)$

answer is B.

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Detailed Solution

$\begin{array}{l} \cos A + \cos B + \cos C = 0 = \sin A + \sin B + \sin C \\ x = c o s A + i \sin A, y = \cos B + i \sin B, z = \cos C + i \sin C \\ x + y + z = 0 \\ x + y = - z \\ {(x + y)}^{3} = - z^{3} \\ x^{3} + y^{3} + 3 x y (x + y) = - z^{3} \\ x^{3} + y^{3} + 3 x y (- z) = - z^{3} \\ x^{3} + y^{3} + z^{3} = 3 x y z \\ {(c i s A)}^{3} + {(c i s B)}^{3} + {(c i s c)}^{3} = 3 c i s A c i s B c i s C \\ c i s 3 A + c i s 3 B + c i s 3 C = 3 c i s (A + B + C) \\ c o m p a r i n g r e a l p a r t s o n b o t h s i d e s \\ \cos 3 A + \cos 3 B + \cos 3 C = 3 \cos (A + B + C) \end{array}$