x^6-64=0 equation

Given the equation
$$x^{6} - 64 = 0$$
Because equation degree is equal to = 6 - contains the even number 6 in the numerator, then
the equation has two real roots.
Get the root 6-th degree of the equation sides:
We get:
$$\sqrt[6]{x^{6}} = \sqrt[6]{64}$$
$$\sqrt[6]{x^{6}} = \left(-1\right) \sqrt[6]{64}$$
or
$$x = 2$$
$$x = -2$$
We get the answer: x = 2
We get the answer: x = -2
or
$$x_{1} = -2$$
$$x_{2} = 2$$

All other 4 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{6} = 64$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{6} e^{6 i p} = 64$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{6 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(6 p \right)} + \cos{\left(6 p \right)} = 1$$
so
$$\cos{\left(6 p \right)} = 1$$
and
$$\sin{\left(6 p \right)} = 0$$
then
$$p = \frac{\pi N}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -2$$
$$z_{2} = 2$$
$$z_{3} = -1 - \sqrt{3} i$$
$$z_{4} = -1 + \sqrt{3} i$$
$$z_{5} = 1 - \sqrt{3} i$$
$$z_{6} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$

The final answer:
$$x_{1} = -2$$
$$x_{2} = 2$$
$$x_{3} = -1 - \sqrt{3} i$$
$$x_{4} = -1 + \sqrt{3} i$$
$$x_{5} = 1 - \sqrt{3} i$$
$$x_{6} = 1 + \sqrt{3} i$$