(x+7)^3=216 equation

Given the equation
$$\left(x + 7\right)^{3} = 216$$
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:
$$\sqrt[3]{\left(x + 7\right)^{3}} = \sqrt[3]{216}$$
or
$$x + 7 = 6$$
Move free summands (without x)
from left part to right part, we given:
$$x = -1$$
We get the answer: x = -1

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x + 7$$
then the equation will be the:
$$z^{3} = 216$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = 216$$
where
$$r = 6$$
- 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 z
Consequently, the solution will be for z:
$$z_{1} = 6$$
$$z_{2} = -3 - 3 \sqrt{3} i$$
$$z_{3} = -3 + 3 \sqrt{3} i$$
do backward replacement
$$z = x + 7$$
$$x = z - 7$$

The final answer:
$$x_{1} = -1$$
$$x_{2} = -10 - 3 \sqrt{3} i$$
$$x_{3} = -10 + 3 \sqrt{3} i$$