x^3+27 equation

Given the equation
$$x^{3} + 27 = 0$$
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:
$$\sqrt[3]{x^{3}} = \sqrt[3]{-27}$$
or
$$x = 3 \sqrt[3]{-1}$$
Expand brackets in the right part

x = -3*1^1/3

We get the answer: x = 3*(-1)^(1/3)

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = -27$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -27$$
where
$$r = 3$$
- 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:
$$z_{1} = -3$$
$$z_{2} = \frac{3}{2} - \frac{3 \sqrt{3} i}{2}$$
$$z_{3} = \frac{3}{2} + \frac{3 \sqrt{3} i}{2}$$
do backward replacement
$$z = x$$
$$x = z$$

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