jmc

We have that $$(-1+i\sqrt{3}{)}^2=(-1+i\sqrt{3})(-1+i\sqrt{3})=1-2i\sqrt{3}-3=-2-2i\sqrt{3},$$and $$(-1+i\sqrt{3}{)}^3=(-1+i\sqrt{3})(-2-2i\sqrt{3})=2+2i\sqrt{3}-2i\sqrt{3}+6=8,$$so $(-1+i\sqrt{3}{)}^6=64.$ Then $${\left(\frac{-1+i\sqrt{3}}{2}\right)}^6=\frac{64}{2^6}=1.$$Similarly, $${\left(\frac{-1-i\sqrt{3}}{2}\right)}^6=\frac{64}{2^6}=1,$$so the expression is equal to $\boxed{2}.$