Differences in implementation of Cardano's method [closed]

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Given a cubic $$ax^3+bx^2+cx+d=0$$ You can divide by $a$, and replace $x$ with $w - \frac{b}{3a}$ to center the cubic and remove the quadratic term

$$ \begin{aligned} ax^3+bx^2+cx+d &= 0 \\ x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} &= 0 \\ (w-\frac{b}{3a})^3+\frac{b}{a}(w - \frac{b}{3a})^2 + \frac{c}{a}(w-\frac{b}{3a})+\frac{d}{a}&=0 \\ w^3-\frac{b}{a}w^2+\frac{b^2}{3a^2}w-\frac{b^3}{27a^3} + \frac{b}{a}w^2-\frac{2b^2}{3a^2}w+\frac{b^3}{9a^3}+\frac{c}{a}w-\frac{bc}{3a^2}+\frac{d}{a}&=0 \\ w^3-\frac{b^2}{3a^2}w+\frac{c}{a}w+\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}&=0 \end{aligned} $$

For the sake of simplicity, let $\frac{c}{a} - \frac{b^2}{3a^2}=p$, and let $\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a} = q$. This would reduce the cubic to a depressed form like so:

$$w^3+pw+q=0$$

From here, I've seen two different ways to solve for $w$. For this, I'll provide an example problem: $w^3 - \frac{1}{3}w - \frac{2}{27} = 0$ (note: $w = x + \frac{4}{3}$).

Method 1: Set $w = u + v$, which would look like this:

$$ \begin{aligned} u^3+3uv(u + v) + v^3+(u + v)p + q= 0 \\ u^3 + v^3 + (3uv + p)(u + v) + q = 0 \end{aligned} $$

This is where my first, smaller question, lies: why set 3ドルuv + p$ to 0? This part does not intuitively make sense to me, however I will proceed through it.

Set 3ドルuv + p = 0$, which would grant us our system of equations:

$$ \begin{aligned} u^3v^3=\frac{-p^3}{27} && && u^3 + v^3 = -q \end{aligned} $$

In our case, $u^3v^3 = \frac{1}{729}$, $u^3 + v^3 = \frac{2}{27}$. Making a quadratic for $u^3$ gives us this:

$$u^6 + u^3q - \frac{p^3}{27} = 0$$

Solving for $u^3$:

$$u^3 = \frac{-q\pm \sqrt{q^2 + \frac{4p^3}{27}}}{2}$$

Now, this brings forth my second question. Which root should I use? We're given two answers for $u^3$, and I am unsure of when to use the positive and negative root. For the example question, the result is this (which doesn't require me to pick sign):

$$u^3 = \frac{\frac{2}{27} \pm \sqrt{\frac{4}{729} - \frac{4}{729}}}{2} = \frac{1}{27}$$

Since $u^3 = \frac{1}{27}$, $v^3 = \frac{1}{27}$ as well, and $w = u + v = \boxed{\frac{2}{3}}$, which is one of the roots of the depressed cubic.

Method 2: Set $w + y = z$, similar to completing the square.

Similar to completing the square, move the constant to one side:

$$w^3 +pw = -q$$

We are assuming that $w^3 + pw$ combined is an incomplete cube, and as such we are going to add $y^3$ to it to make it a full cube of side lengths $z$. To do that, however, we must assume that $pw$ will cover the middle terms of the cubic:

$$pw = 3w^2y + 3wy^2 = 3wy(w + y) = 3wyz$$

This means $p = 3yz$ (meaning $z = \frac{p}{3y}$), and we can add our $y^3$ to the depressed cubic:

$$ \begin{aligned} w^3 + pw + y^3 &= -q + y^3 \\ z^3 &= -q + y^3 \\ \frac{p^3}{27y^3} &= -q + y^3 \\ \frac{p^3}{27} &= -qy^3 + y^6 \end{aligned} $$

Which ends up in another quadratic with respect to $y^3$, and $y^3$ becomes this:

$$ y^3 = \frac{q \pm \sqrt{q^2 + \frac{4p^3}{27}}}{2} $$

For our example, $y^3$ is below:

$$\frac{-\frac{2}{27} \pm \sqrt{\frac{4}{729} - \frac{4}{729} }}{2} = -\frac{1}{27}$$

Meaning $y = -\frac{1}{3} = p$, meaning $z = \frac{1}{3}$, and $w = \boxed{\frac{2}{3}}$, which is the same root found previously.

Now my questions are:

• For Method 1, why set 3ドルuv + p = 0$?
• For Method 1 & 2, which root should you pick in solving the quadratic?
• Why is Method 1 more popular?
• Can you force a specific root? For example, the depressed cubic I gave has a double root at $-\frac{1}{3}$, how would you go about finding that (without dividing the cubic)?

• Jean Marie: Why should this question be closed ? It is very well written and attests work !
• mezzoctane: See also Motivation of Vieta's transformation and Cubic equation#Vieta's substitution

Answer 1

This is where my first, smaller question, lies: why set 3ドルuv + p$ to 0? This part does not intuitively make sense to me, however I will proceed through it.

In the Cardano's method, we had the form $w^3 + p\cdot w + q = 0$ and set $w = u + v$, in order to obtain an "extra degree of freedom":

$$ \begin{aligned} \left(u + v\right)^3 + p\cdot\left(u + v\right) + q = 0 \\ u^3 + 3\cdot u^2\cdot v + 3\cdot u\cdot v^2 + v^3 + p\cdot\left(u + v\right) + q = 0 \\ u^3 + v^3 + 3\cdot u\cdot v\cdot \left(u + v\right) + p\cdot\left(u + v\right) + q = 0 \\ u^3 + v^3 + \left(3\cdot u\cdot v + p\right)\cdot\left(u + v\right) + q = 0 \end{aligned} $$

Now we use the "degree of freedom" by fixing $u$, $v$ so that 3ドル\cdot u\cdot v+p = 0$, in order to get rid of the inconvenient term $\left(3\cdot u\cdot v + p\right)\cdot\left(u + v\right)$ and to reach a quadratic equation for $u^3$:

$$ \begin{aligned} u^3 + v^3 + q = 0 \\ u^3 + \left(-\frac{p}{3\cdot u}\right)^3 + q = 0 \\ u^3 - \frac{p^3}{27\cdot u^3} + q = 0 \\ u^6 + q\cdot u^3 - \frac{p^3}{27} = 0 \\ u^3 = \frac{-q\pm\sqrt{q^2+\frac{4\cdot p^3}{27}}}{2} \\ u^3 = -\frac{q}{2}\pm\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} \end{aligned} $$

Which root should I use? We're given two answers for $u^3$, and I am unsure of when to use the positive and negative root.

The system that led to the quadratic equation was symmetric; the two solutions of the quadratic equation are $u^3$ and $v^3$, and choosing the sign will simply exchange them. We can choose

$$ \begin{aligned} \delta^2 = \Delta = \frac{q^2}{4}+\frac{p^3}{27} \\ u^3 = -\frac{q}{2}+\delta \\ v^3 = -\frac{q}{2}-\delta \end{aligned} $$

If $\Delta > 0$ then there are two possible real choices of $\delta$; the choice is arbitrary but it must be the same in expressions for $u^3$, $v^3$. If $\Delta = 0$ then $u^3 = v^3 = -\frac{q}{2}$. If $\Delta < 0$ then there are two possible complex choices of $\delta$; again, the choice is arbitrary but it must be the same in expressions for $u^3$, $v^3$.

Now, unless $u^3 = 0$, there are three different complex choices of $u$. Likewise, unless $v^3 = 0$, there are three possible complex choices of $v$. However, these choices are not independent because we know 3ドル\cdot u\cdot v+p = 0$; once $u = \sqrt[3]{u^{3}}$ is chosen, $v = -\frac{p}{3\cdot u}$. Three solutions $\left(u, v\right)$ give three solutions $w=u+v$ of the cubic $w^3 + p\cdot w + q = 0$.

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See also questions Understanding Cardano's method, Understanding Cardano's Formula

• LuckElixir: I understand the reason to set 3ドルuv + p = 0,ドル thank you very much for that. For the choice of roots: would the same symmetric situation arise with the second method? Also, why is the first method preferred to find the root, as the second method only assumes that $pw$ covers the middle parts of the cube?