Find all the solutions of the equation 216x^3+2197=0?

To solve the binomial equation, we'll apply the identity
of the sum of cubes:

a^3 + b^3 = (a+b)(a^2 - ab +
b^2)

a^3 =216x^3

a =
6x

b^3 = 13^3

b =
13

216x^3+2197 = (6x+13)(36x^2 - 78x +
169)

If 216x^3+2197 = 0, then (6x+13)(36x^2 - 78x + 169) =
0

If a product is zero, then each factor could be
zero.

6x+13 = 0

We'll
subtract 13 both sides:

6x =
-13

x1 = -13/6

36x^2 - 78x +
169 = 0

We'll apply the quadratic
formula:

x2 = [78 +
sqrt(6084-24336)]/72

x2 =
(78+i*sqrt18252)/72

We'll factorize by 78 the
numerator:

x2 = (78+78i*sqrt
3)/72

x2 = 13(1+isqrt3)/12

x3
= 13(1-isqrt3)/12

The roots of the equation
are: {-13/6 , 13(1+isqrt3)/12, 13(1-isqrt3)/12}.