# Working with casus irreducibilis

I read about casus irreducibilis here. As an example of casus irreducibilis, it says we can factor $x^3 - 15x - 4$ to find $4$ as a root and it also has two other real roots. Using Cardano's method we find $\sqrt{2 + 11i} + \sqrt{2 - 11i}$ as one of the roots. In fact $4 = \sqrt{2 + 11i} + \sqrt{2 - 11i}$ which can be confirmed by noting $(2 + i)^3 = 2 + 11i$ and $(2 - i)^3 = 2 - 11i$.

My question is, using Cardano's method for casus irreducibilis, is there any way to find out what the root actually is (without imaginary numbers) especially if it is a rational number. Is there a way other than guessing the explicit rational number beforehand (like in the example above). The only way I know to calculate the cube root is using Euler's equation but that will require you to know $\cos\left(\frac{\theta}{3}\right)$ and $\sin\left(\frac{\theta}{3}\right)$ but to write that in terms of $\cos(\theta)$ and $\sin(\theta)$ requires you to solve a cubic which starts an unending cycle (where $\theta = \tan^{-1}\left(\frac{b}{a}\right)$ for the complex number $a + bi$).

• The reason for the name is that there seems (in various senses) to be no way to avoid complex numbers. This, more than a raw desire to take square roots of negative numbers, made the complex numbers seem legitimate/necessary. – paul garrett Aug 13 '13 at 0:35
• For integers coefficients, the Rational Roots Theorem seems like the reasonable approach. – André Nicolas Aug 13 '13 at 0:37
• Are we sure this is an example of casus irreducibilis? I mean the roots are $4$, $-2-\sqrt{3}$ and $-2+\sqrt{3}$. I think $x^3 - 3 x + 1 =0$ is a better example. – Emmet Feb 23 '16 at 8:49

## 1 Answer

I don't know if this has any advantage over using the rational root theorem directly, but here is a way you can find the representation if you only have the cube root representation (but it basically finds a polynomial representation and then uses the rational root theorem).

Suppose you don't have the full polynomial but only the solution given by Cardano's formula: $$x =\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$$ (where $b$ is possibly negative: I won't use imaginary numbers explicitly)

I suppose that $x$ is an integer for now. I will explain the rational case later (it works almost exactly the same). Take the cube, and simplify to $$x^3 = 2a + 3x\sqrt{a^2 - b}$$

A first test is to see if $a^2 - b$ equals a cube. If it doesn't, it's impossible to represent the value as an integer. Otherwise, we define $u := \sqrt{a^2 - b}$ (please note that the value of $u$ is known).

$$x^3 = 2a + 3xu$$ $$x^3 - 3xu - 2a = 0$$

You can now use the rational root theorem, and try plugging in divisors of $2a$ for $x$ to see if you found a root. If you find an equality ($x(x^2 + 3) = 2a$), you've found an integer $x$ satisfying $x = \sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$, and otherwise $\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}$ is not an integer.

I think the same approach works for rational instead of integer solutions. The only thing is that you should factor out the denominator at the start so you can still use the rational root theorem.

So, when we want to find the value for $$x = \sqrt{\frac{7}{25}+i\sqrt{\frac{972}{15625}}} + \sqrt{\frac{7}{25}-i\sqrt{\frac{972}{15625}}}$$ factor out the denominator (it is $5$), and bring it to the left side: $$5x = \sqrt{35+i\sqrt{972}} + \sqrt{35-i\sqrt{972}}$$ Substitute $y := 5x$. Continue like we would as explained in the case that $a$, $b$ are integers. We find $u = \sqrt{35^2 - (-972)} = \sqrt{2197} = 13$

The next step is to try divisors of $2a = 70$ in the equation $f(y) := y^3 - 39y - 70 = 0$. We try $y = 1, 2, 5, 7, 35, 70$

$f(1) = -108, f(2) = -140, f(5) = -140, f(7) = 0$

So $$y = \sqrt{35+i\sqrt{972}} + \sqrt{35-i\sqrt{972}} = 7$$ $$x = \sqrt{\frac{7}{25}+i\sqrt{\frac{972}{15625}}} + \sqrt{\frac{7}{25}-i\sqrt{\frac{972}{15625}}} = \frac{y}{5} = \frac{7}{5}$$

You can also use this method to determine if $x =\sqrt{a\pm\sqrt{b}}$ has a respresentation in the form $\frac{c\pm\sqrt{d}}{2}$, since (after applying this method) you have: $$p_+p_- = u = \sqrt{a^2-b}$$ $$p_+ + p_- = x$$

So $p_- = \frac{u}{p_+}$ and if we substitute this in the second equation we get a quadratic equation with $p_+, p_-$ as roots. But again: it's probably easier to just use the rational root theorem on the original cubic equation.

I got started with this by this blog post on Euler, Erdős, however, in this post, the author assumes $b^2 - a = -1$, which makes the equations more elegant. However, this is not the most general case, so the author loses some generality. In the comments I found a link to "Cardan Polynomials and the Reduction of Radicals" by Thomas J. Osler. It looks like it goes in depth much more (I think he shows how to do something similar for radicals of even higher degree than 3), but I have only skimmed the paper.