Expand the expression \((1 + i\sqrt{3})^6\) using De Moivre's theorem.

Step 1 ：De Moivre's theorem states that \((cos(\theta) + isin(\theta))^n = cos(n\theta) + isin(n\theta)\). We'll use this to simplify the expression.

Step 2 ：First, we convert the complex number \(1 + i\sqrt{3}\) into polar form. The magnitude \(r\) is \(\sqrt{1^2 + (\sqrt{3})^2} = 2\). The angle \(\theta\) is \(\arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}\). So \(1 + i\sqrt{3} = 2(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))\).

Step 3 ：Now we can apply De Moivre's theorem. \((1 + i\sqrt{3})^6 = (2(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3})))^6 = 2^6(cos(6\cdot\frac{\pi}{3}) + isin(6\cdot\frac{\pi}{3})) = 64(cos(2\pi) + isin(2\pi))\).

Step 4 ：Finally, we convert back to rectangular form. Since \(cos(2\pi) = 1\) and \(sin(2\pi) = 0\), our final answer is \(64(1 + i\cdot 0) = 64\).