z^3=8 equation

Given the equation
$$z^{3} = 8$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{z^{3}} = \sqrt[3]{8}$$
or
$$z = 2$$
We get the answer: z = 2

All other 2 root(s) is the complex numbers.
do replacement:
$$w = z$$
then the equation will be the:
$$w^{3} = 8$$
Any complex number can presented so:
$$w = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = 1$$
so
$$\cos{\left(3 p \right)} = 1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for w
Consequently, the solution will be for w:
$$w_{1} = 2$$
$$w_{2} = -1 - \sqrt{3} i$$
$$w_{3} = -1 + \sqrt{3} i$$
do backward replacement
$$w = z$$
$$z = w$$

The final answer:
$$z_{1} = 2$$
$$z_{2} = -1 - \sqrt{3} i$$
$$z_{3} = -1 + \sqrt{3} i$$