Cardano's Formula

Theorem
Let $P$ be the cubic equation:
 * $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Then $P$ has solutions:
 * $x_1 = S + T - \dfrac b {3 a}$
 * $x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
 * $x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

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

where:
 * $Q = \dfrac {3 a c - b^2} {9 a^2}$
 * $R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

The expression $D = Q^3 + R^2$ is called the discriminant of the equation.

Real Coefficients
Let $a, b, c, d \in \R$.

Then:

Proof
First the cubic is depressed, by using the Tschirnhaus Transformation:
 * $x \to x + \dfrac b {3 a}$:

Now let:
 * $y = x + \dfrac b {3 a}, Q = \dfrac {3 a c - b^2} {9 a^2}, R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

Thus we have obtained the depressed cubic $y^3 + 3 Q y - 2 R = 0$.

Now let $y = u + v$ where $u v = -Q$.

Then:

Thus the resolvent equation is obtained.

This resolvent is seen to be a quadratic in $u^3$.

From Solution to Quadratic Equation:


 * $u^3 = \dfrac {2 R \pm \sqrt {4 Q^3 + 4 R^2}} 2 = R \pm \sqrt {Q^3 + R^2}$

We have from above $u v = -Q$ and hence $v^3 = -\dfrac {Q^3} {u^3}$.

Let us try taking the positive root: $u^3 = R + \sqrt {Q^3 + R^2}$.

Then:

The same sort of thing happens if you start with $u^3 = R - \sqrt {Q^3 + R^2}$: we get $v^3 = R + \sqrt {Q^3 + R^2}$.

Thus we see that taking either square root we arrive at the same solution.


 * $u^3 = R + \sqrt {Q^3 + R^2}$
 * $v^3 = R - \sqrt {Q^3 + R^2}$
 * $v^3 = R - \sqrt {Q^3 + R^2}$

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

From Roots of Complex Number, we have the three cube roots of $u^3$ and $v^3$:


 * $u = \begin{cases}

& S \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & S \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & S \\ \end{cases}$


 * $v = \begin{cases}

& T \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & T \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & T \\ \end{cases}$

Because of our constraint $u v = -Q$, there are only three combinations of these which are possible such that $y = u + v$:


 * $ y = \begin{cases}

& S + T \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T} \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T} \\ \end{cases}$

As $y = x + \dfrac b {3a}$, it follows that the three roots are therefore:


 * $(1): \quad x_1 = S + T - \dfrac b {3 a}$
 * $(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
 * $(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

Also see

 * Tartaglia's Poem