Theorem: Cardano's Formula

Let $P$ be the cubic equation:
:$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
Then $P$ has solutions:
{{begin-eqn}}
{{eqn | l = x_1
      | r = S + T - \dfrac b {3 a}
}}
{{eqn | l = x_2
      | r = -\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}
}}
{{eqn | l = x_3
      | r = -\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = S
      | r = \sqrt [3] {R + \sqrt {Q^3 + R^2} }
}}
{{eqn | l = T
      | r = \sqrt [3] {R - \sqrt {Q^3 + R^2} }
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = Q
      | r = \dfrac {3 a c - b^2} {9 a^2}
}}
{{eqn | l = R
      | r = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}
}}
{{end-eqn}}

Proof:
First the cubic is depressed, by using the Tschirnhaus Transformation:
:$x \to x + \dfrac b {3 a}$:
{{begin-eqn}}
{{eqn | l = a x^3 + b x^2 + c x + d
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \frac b {3 a} }^3 - 3 \frac b {3 a} x^2 - 3 \frac {b^2} {9 a^2} x - \frac {b^3} {27 a^3} + \frac b a x^2 + \frac c a x + \frac d a
      | r = 0
      | c = completing the cube: $\paren {a + b}^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3$
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } x + \paren {\frac {27 a^2 d - b^3} {27 a^3} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } \paren {x + \frac b {3 a} } - \frac b {3 a} \paren {\frac {3 a c - b^2} {3 a^2} } + \paren {\frac {27 a^2 d - b^3} {27 a^3} }
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x + \frac b {3 a} }^3 + \paren {\frac {3 a c - b^2} {3 a^2} } \paren {x + \frac b {3 a} } + \paren {\frac {27 a^2 d + 2 b^3 - 9 a b c} {27 a^3} }
      | r = 0
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \paren {u + v}^3 + 3 Q \paren {u + v} - 2 R
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u^3 + 3 u^2 v + 3 u v^2 + v^3 + 3 Q \paren {u + v} - 2 R
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u^3 - 3 Q \paren {u + v} + 3 Q \paren {u + v} + v^3 - 2 R
      | r = 0
      | c = as $u v = -Q$
}}
{{eqn | ll= \leadsto
      | l = u^3 + v^3 - 2 R
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u^3 - \frac {Q^3} {u^3} - 2 R
      | r = 0
      | c = substituting for $T$ from $u v = -Q$
}}
{{eqn | ll= \leadsto
      | l = u^6 - 2 R u^3 - Q^3
      | r = 0
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = v^3
      | r = \frac {-Q^3} {R + \sqrt {Q^3 + R^2} }
      | c = 
}}
{{eqn | r = \frac {-Q^3 \paren {R - \sqrt {Q^3 + R^2} } } {R^2 - \paren {Q^3 + R^2} }
      | c = 
}}
{{eqn | r = R - \sqrt {Q^3 + R^2}
      | c = 
}}
{{end-eqn}}
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.
{{WLOG}}:
{{begin-eqn}}
{{eqn | l = u^3
      | r = R + \sqrt {Q^3 + R^2}
}}
{{eqn | l = v^3
      | r = R - \sqrt {Q^3 + R^2}
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = S
      | r = \sqrt [3] {R + \sqrt {Q^3 + R^2} }
}}
{{eqn | l = T
      | r = \sqrt [3] {R - \sqrt {Q^3 + R^2} }
}}
{{end-eqn}}
From Roots of Complex Number, we have the three cube roots of $u^3$ and $v^3$:
:<nowiki>$u = \begin{cases}
& S \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & S \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & S \\ 
\end{cases}$</nowiki>
:<nowiki>$v = \begin{cases}
& T \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & T \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & T \\ 
\end{cases}$</nowiki>
Because of our constraint $u v = -Q$, there are only three combinations of these which are possible such that $y = u + v$:
:<nowiki>$y = \begin{cases}
& S + T \\
\paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T} \\ 
\paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T} \\ 
\end{cases}$</nowiki>
As $y = x + \dfrac b {3 a}$, it follows that the three roots are therefore:
{{begin-eqn}}
{{eqn | n = 1
      | l = x_1
      | r = S + T - \dfrac b {3 a}
}}
{{eqn | n = 2
      | l = x_2
      | r = -\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}
}}
{{eqn | n = 3
      | l = x_3
      | r = -\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}
}}
{{end-eqn}}
{{qed}}