$$\\\dfrac{(2-2i)^6(\sqrt{3}+1)^3}{(\sqrt{3}+i)^4}\\\\\\
\dfrac{(2-2i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(\sqrt{3}+i)^4(\sqrt{3}-i)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{\{(\sqrt{3}+i)(\sqrt{3}-i)\}^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(3--1)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(4)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(2)^8}\\\\\\$$
$$\\\dfrac{(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{4}\\\\\\$$
The rest is just painful expansion. Maybe you do not need to go any further.
(1-i)^6 will work out the easiest. Use binomial expansions.
If you need help expanding I suggest you ask for it. :)
$$\\\dfrac{(2-2i)^6(\sqrt{3}+1)^3}{(\sqrt{3}+i)^4}\\\\\\
\dfrac{(2-2i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(\sqrt{3}+i)^4(\sqrt{3}-i)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{\{(\sqrt{3}+i)(\sqrt{3}-i)\}^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(3--1)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(4)^4}\\\\\\
\dfrac{2^6(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{(2)^8}\\\\\\$$
$$\\\dfrac{(1-i)^6(\sqrt{3}+1)^3(\sqrt{3}-i)^4}{4}\\\\\\$$
The rest is just painful expansion. Maybe you do not need to go any further.
(1-i)^6 will work out the easiest. Use binomial expansions.
If you need help expanding I suggest you ask for it. :)