Typical Omega

Let $a=-1$ and $b=\sqrt{-3}$ . Then we have:

$\begin{aligned} (a+b)^4 + (a-b)^4 & = a^4+4a^3b + 6a^2b^2 + 4ab^3 + b^4 + a^4-4a^3b + 6a^2b^2 - 4ab^3 + b^4 \\ & = 2(a^4 + 6a^2b^2 + b^4) \\ & = 2((-1)^4 + 6(-1)^2(\sqrt{-3})^2 + (\sqrt{-3})^4) \\ & = 2(1+6(-3)+9) \\ & = \boxed{-16} \end{aligned}$

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Alternative Solution

$\begin{aligned} (-1+\sqrt{-3})^4 + (-1-\sqrt{-3})^4 & = (-1+i\sqrt{3})^4 + (-1-i\sqrt{3})^4 \\ & = 2^4\left(-\frac 12 + i \frac {\sqrt 3}2 \right)^4 + 2^4\left(-\frac 12 - i \frac {\sqrt 3}2 \right)^4 \\ & = 16\left(\cos \frac {2 \pi}3 + i \sin \frac {2 \pi}3 \right)^4 + 16\left(\cos \frac {-2 \pi}3 + i \sin \frac {-2 \pi}3 \right)^4 \\ & = 16\left(e^{i \frac {2 \pi}3} \right)^4 + 16\left(e^{-i \frac {2 \pi}3} \right)^4 \\ & = 16 e^{i \frac {8 \pi}3} + 16 e^{-i \frac {8 \pi}3} \\ & = 16 \left(\cos \frac {8 \pi}3 + i \sin \frac {8 \pi}3 \right) + 16 \left(\cos \frac {-8 \pi}3 + i \sin \frac {-8 \pi}3 \right) \\ & = 16 \left(\cos \frac {8 \pi}3 + i \sin \frac {8 \pi}3 \right) + 16 \left(\cos \frac {8 \pi}3 - i \sin \frac {8 \pi}3 \right) \\ & = 32 \cos \frac {8 \pi}3 \\ & = \boxed{-16} \end{aligned}$

${ (-1+\sqrt { -3 } ) }^{ 4 }\quad +\quad { (-1-\sqrt { -3 } ) }^{ 4 }=\quad { (2.\frac { (-1+\sqrt { -3 } ) }{ 2 } ) }^{ 4 }\quad +\quad { (2.\frac { (-1-\sqrt { -3 } ) }{ 2 } ) }^{ 4 }\quad \\ ={ 2 }^{ 4 }.{ \omega }^{ 4 }\quad +\quad { 2 }^{ 4 }.({ { \omega }^{ 2 }) }^{ 4 }\\ ={ 2 }^{ 4 }\left( \omega +{ \omega }^{ 2 } \right) \\ =16\quad \times \quad (-1)\quad =\quad \boxed { -16 }$