To prove that -----  x³ + p x + q = 0 if ---?

...........................

  ..................................................................

The answer is as follows:

Plug that in for x and see what you get.

a = -q/2
b = sqrt((1/4) * q^2 + (1/27) * p^3)

((a + b)^(1/3) + (a - b)^(1/3))^3 + p * ((a + b)^(1/3) + (a - b)^(1/3)) + q

Let's work on each term.  First, x^3

(a + b)^(3/3) + 3 * (a + b)^(2/3) * (a - b)^(1/3) + 3 * (a + b)^(1/3) * (a - b)^(2/3) + (a - b)^(3/3) =>
a + b + 3 * (a + b)^(1/3) * (a + b)^(1/3) * (a - b)^(1/3) + 3 * (a + b)^(1/3) * (a - b)^(1/3) * (a - b)^(1/3) + a - b =>
2a + 3 * (a + b)^(1/3) * (a - b)^(1/3) * ((a + b)^(1/3) + (a - b)^(1/3)) =>
2a + 3 * ((a + b) * (a - b))^(1/3) * x =>
2a + 3 * (a^2 - b^2)^(1/3) * x =>
2a + 3 * ((1/4) * q^2 - (1/4) * q^2 - (1/27) * p^3)^(1/3) * x =>
2a + 3 * ((-1/27) * p^3)^(1/3) * x =>
2a + 3 * (-1/3) * p^(3/3) * x =>
2a - px

Now we have:

2a - px + px + q =>
2a + q =>
2 * (-q/2) + q =>
-q + q =>
0

0 = 0

Done.