+6248 \(\text{The easiest way to go about this is to convert the number to polar form}\\ -2-2\sqrt{3}i = -2(1+\sqrt{3}i) = \\ -2 \cdot 2\left(\dfrac 1 2 + \dfrac{\sqrt{3}}{2}i\right) = \\ -4\left(\cos\left(\dfrac \pi 3\right) + i \sin\left(\dfrac \pi 3\right)\right) = \\ \large 4 e^{i \pi} \cdot e^{i\frac \pi 3} = \\ \large 4 e^{i \frac{4\pi}{3}}\)
\(\large \left(4 e^{i\frac{4\pi}{3}}\right)^3 = \\ \large 4^3 \cdot e^{3\left(i \frac{4\pi}{3}\right)} = \large \\ 64 e^{i 4 \pi} = \\ 64\)
\(|64|=64\)
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OK here is hopefully the correct answer.
\(\left(-2\right)^3-3\left(-2\right)^2\cdot \:2\sqrt{3}i+3\left(-2\right)\left(2\sqrt{3}i\right)^2-\left(2\sqrt{3}i\right)^3\)=
\(\left(-2\right)^3-3\left(-2\right)^2\cdot \:2\sqrt{3}i-3\cdot \:2\left(2\sqrt{3}i\right)^2-\left(2\sqrt{3}i\right)^3\)=
\(-8-\left(-2\right)^2\cdot \:3\cdot \:2\sqrt{3}i-3\cdot \:2\left(2\sqrt{3}i\right)^2-\left(2\sqrt{3}i\right)^3\)=
\(3\left(-2\right)^2\cdot \:2\sqrt{3}i\)=
\(2^2\cdot \:3\cdot \:2\sqrt{3}i\)=
\(24\sqrt{3}i\)=
\(-8+72\)=
\(64\)
Because you are looking for the absulate value the answer is
\(\begin{align*} 64. \end{align*}\)
+141 Another way would be to evaluate \(|-2-2i\sqrt 3|^3\). We have: \(\sqrt{(-2)^2+(-2\sqrt 3)^2}=\sqrt{4+4\cdot 3}=\sqrt{16}=4\implies 4^3=\boxed{64}~~\blacksquare\)
