(x-1)^3=-8 equation

Given the equation
$$\left(x - 1\right)^{3} = -8$$
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:
$$\sqrt[3]{\left(x - 1\right)^{3}} = \sqrt[3]{-8}$$
or
$$x - 1 = 2 \sqrt[3]{-1}$$
Expand brackets in the right part

-1 + x = -2*1^1/3

Move free summands (without x)
from left part to right part, we given:
$$x = 1 + 2 \sqrt[3]{-1}$$
We get the answer: x = 1 + 2*(-1)^(1/3)

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x - 1$$
then the equation will be the:
$$z^{3} = -8$$
Any complex number can presented so:
$$z = r e^{i p}$$
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
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
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:
$$z_{1} = -2$$
$$z_{2} = 1 - \sqrt{3} i$$
$$z_{3} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x - 1$$
$$x = z + 1$$

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