If $\cos { \alpha } +\cos { \beta } +\cos { \gamma } =\sin { \alpha } +\sin { \beta } +\sin { \gamma } =0$, then the value of $\cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma }$ is

Let $a=\cos { \alpha } +i\sin { \alpha } ,b=\cos { \beta } +i\sin { \beta } ,c=\cos { \gamma } +i\sin { \gamma }$

The, $a+b+c=\left( \cos { \alpha } +\cos { \beta } +\cos { \gamma } \right) +i\left( \sin { \alpha } +\sin { \beta } +\sin { \gamma } \right) =0+i0=0$

$\Rightarrow { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc$

$\Rightarrow \left( \cos { 3\alpha } +i\sin { 3\alpha } \right) +\left( \cos { 3\beta } +i\sin { \beta } \right) +\left( \cos { 3\gamma } +i\sin { 3\gamma } \right)$

$=3\left[ \cos { \left( \alpha +\beta +\gamma \right) } +i\sin { (\alpha +\beta +\gamma ) } \right]$

by equating the real components

$\Rightarrow \cos { 3\alpha } +\cos { 3\beta } +\cos { 3\gamma } =3\cos { \left( \alpha +\beta +\gamma \right) }$