3
$\begingroup$

We can derive roots of depressed cubic using Cardano method:
$x=u+v$ (1)
We then substitute x, group variables:
$u^6+q \cdot u^3- \frac{p^3}{27} = 0$ (2)
or
$w^2+q \cdot w- \frac{p^3}{27} = 0$ (3)

Consider a depressed cubic equation

$(x-5)\cdot(x+1)\cdot(x+4)=0$ (4)
$x^{3} -21 \cdot x-20=0$ (5)
Formula yields solution
$x = \sqrt[3]{10 + \sqrt{-243}} + \sqrt[3]{10 - \sqrt{-243}}$ (6)
Good news: we expressed solution in radicals and WolframAlpha knows that:
$2\cdot\sqrt[3]{10 + \sqrt{-243}} = 5 + i\cdot\sqrt{3}$ (7)

Bad news: I cannot get from (6) to (7) on my own. If one tries to extract cubic root using algebra - we arrive back to cubic equation. Unlike real numbers there is no simple iterative method to extract cubic roots of complex numbers. Trigonometry helps only if Euler formula is known. However, complex numbers were not recognized back then.

What would Cardano do in this case? Did he just give up or were there numeric methods to express the value?

$\endgroup$

2 Answers 2

4
$\begingroup$

It can be proven that, when the cubic equation has three real roots, it is not possible to express them by a function of the coefficients involving only real radicals. So Cardano was stuck in cases like this, unless he already knew the roots for some other reasons.

When the polynomial is irreducible over $\mathbb{Q}$ this was in fact called casus irreducibilis, and its occurrence was the leading reason for the introduction of complex numbers by Bombelli and others.

$\endgroup$
2
$\begingroup$

The simplification of the cube root can be brute forced using a little algebraic number theory. Write $\omega = \frac{-1+\sqrt{-3}}{2}$, so we'll need only to consider integer solutions. Let $\alpha = 10 + 9\sqrt{-3} = 19+18\omega$. Suppose that $\alpha$ is a perfect cube, then $\alpha = (a+b\omega)^3$ for some integers $a$ and $b$. Expanding using the binomial theorem $$ \alpha = a^3+3a^2b\omega + 3ab^2\omega^2+b^3\omega^3 $$ Equating coefficients and simplifying leads to the following two equations $$ \begin{align} a^3-3ab^2+b^3&=19 \\ ab(a-b)&= 6 \end{align} $$ We're fairly lucky here, because it's easy to see that $a=3$ and $b=1$ is a solution. Then a root of the cubic is $(a+b\omega)+(a+b\omega^2)=2a-b=6-1=5$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .