jmc

We know that ${\omega }^3-1=0,$ which factors as $(\omega -1)({\omega }^2+\omega +1)=0.$ Since $\omega$ is not real, ${\omega }^2+\omega +1=0.$

Then $$(1-\omega +{\omega }^2{)}^4+(1+\omega -{\omega }^2{)}^4=(-2\omega {)}^4+(-2{\omega }^2{)}^4=16{\omega }^4+16{\omega }^8.$$Since ${\omega }^3=1,$ this reduces to $16\omega +16{\omega }^2=16({\omega }^2+\omega )=\boxed{-16}.$