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Concerning Cardano's solution formula for the depressed cubic equation \begin{equation*} x^3 + px + q = 0 \end{equation*} given by \begin{equation*} x=\sqrt[3]{-q/2+\sqrt{q^2/4+p^3/27}}+\sqrt[3]{-q/2-\sqrt{q^2/4+p^3/27}} \end{equation*} it is known to give in some cases tautologous answers leading to a vicious circle.

E.g. the equation \begin{equation} \label{equ} x^3 + 3x - 4 = 0 \end{equation} clearly has, by inspection, the solution $x_1 = 1$ (and so the other solutions $x_{2,3} = -1/2 \pm \sqrt{-5}/2)$, but Cardano's solution formula gives for the positive solution \begin{equation} \label{sol} x=\sqrt[3]{2+\sqrt{5}}-\sqrt[3]{-2+\sqrt{5}}. \end{equation} If one tries to simplify this getting rid of the roots by cubing this, one gets \begin{equation*} \begin{split} x^3 &= 2+\sqrt{5}-(-2+\sqrt{5}) - 3\sqrt[3]{(2+\sqrt{5})(-2+\sqrt{5})}x\\ &= 4 - 3\sqrt[3]{(-4+5)}x \end{split} \end{equation*} ending up with the given equation in a vicious circle: $x$ has just that value needed to make this equation true.

Question: By which algebraic manipulations may one show $x = 1$?

MathCrawler
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2 Answers2

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It may help to denest the roots, as in the vein of the answer I gave here.

So let's look at $\sqrt[3]{2 + \sqrt{5}}$. We seek rational $a,b$ such that $$ 2+\sqrt 5 = (a+b \sqrt 5)^3 = (a^3 + 15 ab^2) + (3 a^2 b + 5 b^3) \sqrt 5 $$ Naturally, comparing the coefficients on each side, we see that the necessary equations to satisfy are $$\begin{cases} a^3 + 15ab^2 = 2 \\ 3a^2 b + 5b^3 = 1 \end{cases}$$ I'm not sure if there's a general way to solve equations of this nature beyond guessing and checking, graphing, or other such types of methods. I guess one could think "it would be nice if $a=b$" and reach the equations $$\begin{cases} 16a^3 = 2 \\ 8a^3 = 1 \end{cases}$$ and hence conclude "easily" that $a=b=1/2$, as with KCd's answer.

EDIT: An alternative method:

I didn't think to read through my post from before, but it leads to a method of solution that feels less arbitrary.

Note that, doubling the second equation, $$ a^3 + 15ab^2 = 2 = 6a^2 b + 10b^3 $$ and so $$ a^3 - 6a^2 b + 15ab^2 - 10b^3 = 0 $$ Divide throughout by $b^3$: $$ \left( \frac a b \right)^3 - 6 \left( \frac a b \right)^2 + 15 \left( \frac a b \right) - 10 = 0 $$ or, with a variable $z$, $$z^3 - 6z^2 + 15z - 10 = 0$$ Find a solution $z$ to this equation via a method of your preference. While it is still a "guess and check" method, the rational root theorem will let you conclude that $z=1$ is a valid root. That is, $$ z = \frac a b = 1 \implies a = b $$ (if this indeed the root we should use; my linked post addresses this a bit). This allows us to reduce to the better set of equations, $$\begin{cases} 16a^3 = 2 \\ 8a^3 = 1 \end{cases}$$ and conclude as in KCd's answer.

One can do the analogous calculation for $\sqrt[3]{-2 + \sqrt 5}$ to denest it, and take the difference to conclude.

PrincessEev
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First of all, do not expect that you must always be able to find a "nice" solution via Cardano's formula. Ultimately the meaning of the formula is that there is an expression for the roots in terms of iterated square roots and cube roots, not that nice final answers must have nice derivations from Cardadno's formula.

But in your example, something can be done. By omniscience (that's a joke), it turns out that $$ 2 + \sqrt{5} = \left(\frac{1+\sqrt{5}}{2}\right)^3, $$ so $$ \sqrt[3]{2+\sqrt{5}} - \sqrt[3]{-2+\sqrt{5}} = \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = \frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2} = 1. $$

KCd
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