Explanation

(c)
We have, $\cos{\alpha + cos\beta + \cos{\gamma = 0}}$ …(i)
and $\sin{\alpha + \sin{\beta + \sin{\gamma = 0}}}$ …(ii)
Let $a = \cos{\alpha + i\sin\alpha}$;
$b = \cos{\beta + i\sin\beta}$
and $c = \cos{\gamma + i\sin\gamma}$
Therefore, $a + b + c = (\cos{\alpha + cos\beta + \cos{\gamma)}}$
$+ i\ (\sin{\alpha + \sin{\beta + \sin\gamma}})$
$= 0 + i0 = 0\ \ \lbrack$from Eqs.(i)and (ii)]
If $a + b + c = 0,$ then $a^{3} + b^{3} + c^{3} = 3abc$
$\Rightarrow \ \left( \cos{\alpha + i\sin\alpha} \right)^{3} + \left( \cos{\beta + i\sin\beta} \right)^{3} + \left( \cos{\gamma + i\sin\gamma} \right)^{3}$
$= 3(\cos{\alpha + i\sin{\alpha)}}(\cos{\beta + i\sin\beta})(\ \cos{\gamma + i\sin\gamma}$)
$\Rightarrow \ (\cos{3\alpha + i\sin{3\alpha)}} + (\cos{3\beta + i\sin{3\beta)}} + (\cos{3\gamma + i\sin{3\gamma}}$)
$= 3\lbrack\cos{(\alpha + \beta + \gamma) + i\sin{(\alpha + \beta + \gamma)\rbrack\ }}$
$\Rightarrow \ \cos{3\alpha + \cos{3\beta + \cos{3\gamma = 3\cos(\alpha + \beta + \gamma)}}}$