Cardano's Formula/Examples/x^3 - 15x - 4

Example of Use of Cardano's Formula

 * $\ds x^3 - 15x - 4 = 0$

has solutions $x = 4$, $x = -2 + \sqrt 3$ and $x = -2 - \sqrt 3$.

Proof
This is in the form:
 * $a x^3 + b x^2 + c x + d = 0$

where:
 * $a = 1$
 * $b = 0$
 * $c = -15$
 * $d = -4$

From Cardano's Formula:
 * $x = S + T$

where:
 * $S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
 * $T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

where:

and:

Thus:

Using Cardano's Formula/Real Coefficients, we know from
 * $D = Q^3 + R^2 = -121 < 0$

that all roots are real and unequal.

From Roots of Complex Number/Examples/Cube Roots of 2+11i, we have:
 * $\sqrt [3] {\paren {2 + 11 i } } = \set {2 + i, -1 + \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 - \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {-\dfrac 1 2 + \sqrt 3}}$

From Roots of Complex Number/Examples/Cube Roots of 2-11i, we have:
 * $\sqrt [3] {\paren {2 - 11 i } } = \set {2 - i, -1 + \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 + \sqrt 3 }, -1 - \dfrac {\sqrt 3} 2 + i \paren {\dfrac 1 2 - \sqrt 3}}$

We need to investigate $9$ sums and find solutions where the complex part vanishes.

Our $3$ solutions are:

Hence:

The result follows.