Provided: \( z = 2(\cos 60^\circ + i \sin 60^\circ) = 1 + i \sqrt{3} \). Applying binomial expansion: \( z^3 = (1 + i \sqrt{3})^3 \). First, \( z^2 = (1 + i \sqrt{3})^2 = 1 + 2 i \sqrt{3} + (i \sqrt{3})^2 = 1 + 2 i \sqrt{3} - 3 = -2 + 2 i \sqrt{3} \). Next, \( z^3 = z^2 \times z = (-2 + 2 i \sqrt{3})(1 + i \sqrt{3}) = -2 - 2 i \sqrt{3} + 2 i \sqrt{3} + 2 (i \sqrt{3})^2 = -2 + 0 + 2(-3) = -2 - 6 = -8 \). Therefore, \( z^3 = -8 \).