To solve the equation $z^3 = 8$, we can rewrite it as:

$z^3 = 8$

The cube root of 8 is 2, since:

$2^3 = 8$

Thus, one solution is $z = 2$.

However, since we are dealing with a cubic equation, there can be more solutions in the complex plane. To find all solutions, we express 8 in polar form. The number 8 can be written as:

$8 = 8 (\cos 0^\circ + i \sin 0^\circ)$

Using De Moivre's Theorem, the general solution for $z^3 = 8$ is:

$z = 2 \left[ \cos \left( \frac{0^\circ + 360^\circ k}{3} \right) + i \sin \left( \frac{0^\circ + 360^\circ k}{3} \right) \right]$ where $k = 0, 1, 2$.

Now, let's calculate each solution:

1. For $k = 0$: $z_0 = 2 \left[ \cos 0^\circ + i \sin 0^\circ \right] = 2$

2. For $k = 1$: $z_1 = 2 \left[ \cos 120^\circ + i \sin 120^\circ \right] = 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = -1 + i\sqrt{3}$

3. For $k = 2$: $z_2 = 2 \left[ \cos 240^\circ + i \sin 240^\circ \right] = 2 \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -1 - i\sqrt{3}$

Thus, the three solutions to the equation $z^3 = 8$ are:

$z_0 = 2, \quad z_1 = -1 + i\sqrt{3}, \quad z_2 = -1 - i\sqrt{3}$

Would you like a more detailed explanation of De Moivre's Theorem or the complex number approach?