## Cardano's Method

Cardano's method of solving a cubic applies directly to depressed cubics, those of the form $$ x^3 + px + q = 0 $$

We should however note that Cardano's method often gives answers in very unusual form; in particular, the method sometimes expresses real solutions as a function of complex numbers, which can be both unhelpful and disconcerting.

The key idea behind the method is to express our variable, $x$ as the sum of two other variables, $x = u+v$ and then note we can impose certain conditions on $u$ and $v$ as our choice of $u$ and $v$ was free.

## Cardano's Method 1

We shall use Cardano's method of solving a cubic, be warned however, the algebra gets... messy...

We first introduce new variables $u$ and $v$ such that $z =u+v$. Substitution of these variables into our equation gives

$$\begin{align}(u+v)^3 + (u+v) - 10 &=0\\ u^3 + v^3 + 3uv(u+v) + 1(u+v) - 10 &= 0 \\ u^3 + v^3 + (3uv + 1)(u+v) - 10 &= 0 \end{align} $$

Because we are totally free to choose the variables u and v, let us insist that they have the property that $(3uv + 1) = 0$.

Then our variables u,v must now satisfy $$ 3uv = -1,\;u^3 + v^3 = 10 $$

Let us construct a quadratic equation that satisfies these conditions: Consider the quadratic expression $$ \begin{align} (x- u^3)(x - v^3) &= x^2 - (u^3 + v^3)x + (uv)^3 \\ &= x^3 - 10x + \left(-\frac{1}{3}\right)^3 \end{align} $$

This gives rise to the quadratic equation $x^3 - 10x -\frac{1}{27} =0$, the solutions of which are $u^3$ and $v^3$.

In this case, we get $x = \frac{45\pm 26\sqrt{3}}{9}$. At this point, the work starts to get very messy.

Let us take the positive solution, $u^3= \frac{45 +26\sqrt{3}}{9}$, then we get

$$u = \sqrt[3]{\frac{45 +26\sqrt{3}}{9}}$$

with two other options for $u$ possible by multiply by roots of unity.

We note that $3uv = -1$, so $v = \frac{-1}{3u}$ and hence we get our first solution,

$$z = \sqrt[3]{\frac{45 +26\sqrt{3}}{9}} + \frac{-1}{3\sqrt[3]{\frac{45 +26\sqrt{3}}{9}}}$$

This is obviously a completely horrendous surd, but feeding this monstrosity into a calculator reveals a rather surprising outcome: $z = 2$. We could now proceed to calculate the other solutions by considering $\omega u$ and $\omega^2u$, where $\omega = e^{i\frac{2\pi}{3}}$, but I would suggest once one root has been found, just use algebraic long division to find a quadratic that the second two solutions satisfy!

Using algebraic division yields $z^3 + z - 10 =(z-2)(z^2 +2z + 5)$. This in turn means our other two solutions satisfy $z^2 +2z + 5=0$, giving a final pair of conjugate solutions. $z = 1 \pm 2i$

## Cardano's Method 2

Let $x= u + v$, then $$(u+v)^3 - 3(u+v) + 1 = 0 $$

Expansion and simplification gives $$u^3 + v^3 + 3(uv^2 + u^2v) - 3(u+v) + 1 = 0$$

This can be rewritten as $$ (u^3 + v^3 + 1) + 3(uv - 1)(u+v) = 0 $$

If follows that if $u^3 + v^3 = -1$ and $uv =1$ then u,v satisfy our requirements. From this pair of conditions, we construct the quadratic (using $(y - u^3)(y - v^3)=0$) $$ y^2 + y +1 = 0 $$

Hence, without loss of generality, $u^3 = \sqrt[3]{\frac{-1 + \sqrt{3}i}{2}}$, $v^3 = \frac{1}{u^3}$

The curious thing to note here is that $$u^3 = e^{i\frac{2\pi}{3}}$$ Hence, using the polar form of complex numbers, we can move more quickly to a solution $$ u = e^{i(\frac{2\pi}{9} + \frac{2\pi n}{3})} $$ The corresponding value of $v$ is simply the reciprocal; giving

$$v = e^{-i(\frac{2\pi}{9} + \frac{2\pi n}{3})}$$

This in turn means our solutions can be expressed beautifully in trig form! (via de Moivre's theorem)

$$\begin{align}x &= u + v\\ &=e^{i(\frac{2\pi}{9} + \frac{2\pi n}{3})} + e^{-i(\frac{2\pi}{9} + \frac{2\pi n}{3})} \\ &= 2\cos\left(\frac{2\pi}{9} + \frac{2\pi n}{3}\right)\end{align} $$

Where $n = 0,1,2$. Note that all three solutions are real!