I really don't have much of my own attempt to show for this question. All I managed to get to was:
$$z^2(z-3\sqrt{3}i)-3(3z-3\sqrt{3}i)$$
By simply factorising. It is asked in the question to be completed via the method of grouping. I have been starring at this for far too long and don't see any way for this to simplify down nicely. The answer should be $(z-\sqrt{3}i)^3$.
Start by grouping the terms as $$(z^3-2iz\sqrt{3}-3z)+(-iz^2\sqrt{3}-6z+3i\sqrt{3})$$ Then factoring out from the brackets: $$z(z^2-2iz\sqrt{3}-3)-i\sqrt{3}(z^2-2iz\sqrt{3}-3)$$ $$=(z-i\sqrt{3})(z^2-21z\sqrt{3}-3)$$ $$=(z-i\sqrt{3})(z-i\sqrt{3})^2$$
et voila!