$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }=$

We have $\displaystyle \frac { \sqrt { 3 } +i }{ 2 } =\frac { i\sqrt { 3 } +{ i }^{ 2 } }{ 2i }$

(Here we multiplied the numerator and denominator by $(-i)$ on the LHS)

We get: $-i\left( \frac { -1+\sqrt { 3 } i }{ 2 } \right) =i\omega$

Simliarly applying the same idea we get the steps below:

and $\displaystyle \frac { i-\sqrt { 3 } }{ 2 } =\frac { { i }^{ 2 }-i\sqrt { 3 } }{ 2i } =-i\left( \frac { -1-\sqrt { 3 } i }{ 2 } \right) =-i{ \omega }^{ 2 }$

Hence $\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }={ \left( -i\omega \right) }^{ 6 }+{ \left( -i{ \omega }^{ 2 } \right) }^{ 6 }\\ \\ ={ i }^{ 6 }\left( { \omega }^{ 6 }+{ \omega }^{ 12 } \right) =-1\left( 1+1 \right) =-2$