Answer
$4096i$
Work Step by Step
Since, $(2\sqrt 2-2\sqrt 2i)=4(\cos 315^{\circ}+i \sin 315^{\circ})$
and $(2\sqrt 2-2\sqrt 2i)^{6}=4(\cos 315^{\circ}+i \sin 315^{\circ})^{6}$
De Moivre’s Theorem states that when $ r (\cos\theta+i \sin\theta)$ is a complex number, and if $n$ is any real number, then the following relationship holds.
$[ r (\cos\theta+i \sin\theta)]^{n}=[ r^{n} (\cos n\theta+i \sin n\theta)]$
In compact form, this is written
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$4(\cos 315^{\circ}+i \sin 315^{\circ})^{6}=4^{6}(\cos 6\times315^{\circ}+i \sin 6\times315^{\circ})$
$=4096(\cos 1890^{\circ}+i \sin 1890^{\circ})$
$=4096(\cos 90^{\circ}+i \sin 90^{\circ})$
$=[4096(0+i.1)]$
$=4096i$
