jmc

Let $z=\frac{3+i\sqrt{3}}{2}$ and $w=\frac{3-i\sqrt{3}}{2}.$ Then $$z+w=\frac{3+i\sqrt{3}}{2}+\frac{3-i\sqrt{3}}{2}=3$$and $$zw=\frac{3+i\sqrt{3}}{2}\cdot \frac{3-i\sqrt{3}}{2}=\frac{9-3i^2}{4}=\frac{9+3}{4}=3.$$Squaring the equation $z+w=3,$ we get $$z^2+2zw+w^2=9,$$so $z^2+w^2=9-2zw=3.$

Squaring the equation $z^2+w^2=3,$ we get $$z^4+2z^2{w}^2+w^4=9,$$so $z^4+w^4=9-2z^2{w}^2=9-2\cdot 3^2=-9.$

Finally, squaring the equation $z^4+w^4=-9,$ we get $$z^8+2z^4{w}^4+w^8=81,$$so $z^8+w^8=81-2z^4{w}^4=81-2\cdot 3^4=\boxed{-81}.$