I'm trying to understand how to solve cubic equations using Cardano's formula. To test the method, I expand $(x-3)(x+1)(x+2)=x^3-7x-6$. My hope is that the formula will produce the roots $-1,-2,3$. But the formula seems to make a mess of things: I compute that $\frac{q^2}{4}+\frac{p^3}{27}=\frac{100}{27}$, and so the formula gives me the baffling \begin{equation} \sqrt[3]{3+i \sqrt{\frac{100}{27}}}+\sqrt[3]{3-i \sqrt{\frac{100}{27}}}. \end{equation}

I'd like to know if there is a straightforward way one or all of the roots $-1,-2,3$ from this expression. I've asked several people this question, and the usual punchline is that I've produced a proof that this expression is $-1,-2$ or $3$ (depending on the choice of cube root etc.) That is not my goal.

I found a book of Cardano's writings in the library, but it seems some of his writings have been lost. I'm convinced that he and his cohort had some method for doing this. So, does anyone know how to use the cubic formula for real? Specifically, in such a way as to recognize the output as a particular integer/rational number when it is one?



This is the casus irreducibilis, first discussed in detail by Bombelli. We end up unavoidably needing to travel through the complex numbers to end up with real roots! This is important historically, since it was the first time that one needed to treat complex non-real numbers seriously. We do not need to worry about non-real numbers when solving quadratics, since after all we can say that there are no roots. But that is not the case here, since undeniably there are real roots.

When you are trying to find the cube roots of your complex expression, you can assume that a cube root of your first expression is $a+bi$. Cube this, and you will get some messy equations. But you can (in this case) "spot" a root, and then you are finished. But that is cheating, it is due entirely to the fact that the roots of the original cubic are "nice."

One workaround of sorts is to use the trigonometric solution to the cubic, which is uses the trigonometric identity $\cos 3\theta=4\cos^3\theta-3\cos\theta$. You may want to look at the Wikipedia article on the cubic, which is reasonably thorough.

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    $\begingroup$ I looked at Wikipedia; it didn't really answer my question. I'm trying to find a way to use Cardano's formula. It seems it is only of theoretical interest? $\endgroup$ – Steven Spallone May 16 '12 at 1:28
  • $\begingroup$ Indeed, you are right, except we can find an expression for the required cube root by using De Moivre's Theorem. But that takes us again to transcendental functions, just like the trigonometric solution. $\endgroup$ – André Nicolas May 16 '12 at 1:55

I know it has been a lot of time since you asked the question, but it is reasonable to assume that no one has answered you. Ok, here we go. When trying to reduce radicals in general, you try to find values for a, b $\in \mathbb{Q}$ such that $$\sqrt[n]{A \pm B\sqrt[m]{C}} = a \pm b\sqrt[m]{C}.$$Notice that $\sqrt[m]{C}$ stays the same. What we are going to change the original expression to such form: $$\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = \sqrt[3]{3 \pm \frac{10}{3}\frac{i}{\sqrt{3}}} = \sqrt[3]{3 \pm \frac{10}{9}(i\sqrt{3})}$$ We will now try to find values for a, b $\in \mathbb{Q}$ such that $$\sqrt[3]{3 \pm \frac{10}{9}(i\sqrt{3})} = a \pm b(i\sqrt{3}).$$Cubing both sides:$$3 \pm \frac{10}{9}(i\sqrt{3}) = (a^3 - 9ab^2) \pm (3a^2b - 3b^3)(i\sqrt{3})$$ Since a and b are rational numbers, then the previous equation is only true if:$$\begin{cases}a^3 - 9ab^2 = 3\\3a^2b - 3b^3 = \frac{10}{9}\end{cases}\Rightarrow\begin{cases}a^3 - 9ab^2 = 3\\27a^2b - 27b^3 = 10\end{cases}$$To solve that system of equations we are going to multiply one equation by the other:$$(3)(27a^2b - 27b^3) = (10)(a^3 - 9ab^2)\\81a^2b - 81b^3 = 10a^3 - 90 ab^2\\10a^3 - 81a^2b - 90ab^2 + 81b^3 = 0$$Dividing both sides by $b^3$:$$10(\frac ab)^3 - 81(\frac ab)^2 - 90(\frac ab) + 81 = 0$$We are now going to define x = a/b. Now, since a and b are rational numbers, x is also rational. Using the rational root theorem, we can find the solutions for our cubic equation:$$10x^3 - 81x^2 - 90x + 81 = 0\\x = -{3 \over 2} \lor x = {3 \over 5} \lor x = 9$$If x = -3/2, then a = -1 and b = 2/3.

If x = 3/5, then a = -1/2 and b = -5/6.

If x = 9, then a = 3/2 and b = 1/6.

Therefore:$$\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -1 \pm i\frac{2\sqrt{3}}{3}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -\frac{1}{2} \mp i\frac{5\sqrt{3}}{6}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = \frac{3}{2} \pm i\frac{\sqrt{3}}{6}$$And there you have it. I learned this method here, by the way, after comming across the same problem you did.

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    $\begingroup$ What if there is more than two terms under the cube root? $\endgroup$ – Zzz Nov 9 '16 at 15:49
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    $\begingroup$ Thanks Bruno! However, this seems circular, since you might as well use the rational root test to solve the original cubic, and not bother with nested radicals at all. So why bother with Cardano's formula? $\endgroup$ – Steven Spallone Apr 24 '17 at 16:30
  • $\begingroup$ @StevenSpallone Yes, it feels really pointless, doesn't it? But the thing is that we were able to prove that $$ \sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -1 \pm i\frac{2\sqrt{3}}{3}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = -\frac{1}{2} \mp i\frac{5\sqrt{3}}{6}\\\sqrt[3]{3 \pm i \sqrt{\frac{100}{27}}} = \frac{3}{2} \pm i\frac{\sqrt{3}}{6} $$ $\endgroup$ – Bruno Monteiro Apr 24 '17 at 17:43

Cardano's formulas work like that. When a polynomial with real coefficients has three distinct real roots, the formulas give you two of the roots with complex numbers as intermediate steps. In your case, in fact: $$ \sqrt[3]{3+i \sqrt{\frac{100}{27}}} = \frac{3}{2}+i\frac{\sqrt{3}}{6}, \qquad \sqrt[3]{3-i \sqrt{\frac{100}{27}}} = \frac{3}{2}-i\frac{\sqrt{3}}{6}, $$ so their sum is $3$.

  • $\begingroup$ Okay, so here we are. How do you see "in fact..." without knowing already that -1,-2,3 are roots? To me, these facts seem more difficult to establish! $\endgroup$ – Steven Spallone May 16 '12 at 1:35
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    $\begingroup$ Cardano's formulas work like that. When a polynomial with real coefficients has three distinct real roots, the formulas give you two of the roots with complex numbers as intermediate steps. In your case, in fact: $$ \sqrt[3]{3+i \sqrt{\frac{100}{27}}} = \frac{3}{2}+i\frac{\sqrt{3}}{6}, \qquad \sqrt[3]{3-i \sqrt{\frac{100}{27}}} = \frac{3}{2}-i\frac{\sqrt{3}}{6}, $$ so their sum is $3$. How on earth do you figure out things like $\sqrt[3]{3+i \sqrt{\frac{100}{27}}} = \frac{3}{2}+i\frac{\sqrt{3}}{6}$? $\endgroup$ – user42091 Sep 21 '12 at 16:44

It is possible to prove that if $a$ and $b$ are relatively prime integers such that $a$ is not divisile by $3,$ then the cubic $x^{3}+ax+b$ has all its roots integers if and only if $4a^{3}+27b^{2}= -c^{2}$ for some integer $c.$ If $a$ is divisible by $3,$ but $a$ and $b$ are still relatively prime, then the situation is more complicaed, and there are three integer roots if and only if $4a^{3}+27b^{2} = -729c^{2}$ for some integer $c.$ In the first case, there are integers $s$ and $t$ such that $-a = s^{2}+t^{2}-st,$ $b = st(s-t)$, and the roots are $-s,t$ and $s-t$. In the second case, there are integers $s$ and $t$ such that $-a = 3(s^{2}+t^{2}-st)$ and the roots are $s+t, t-2s$ and $s-2t$. The only way I know to prove this myself is using the fact that the Eisenstein integers are a unique factorization domain. The ring of Eisenstein integers is $\mathbb{Z}[\omega]$, where $\omega$ is a complex primitive cube root of unity.

  • $\begingroup$ Thanks for the info, though I still don't see a straightforward way to manipulate the expression to get the answer. $\endgroup$ – Steven Spallone May 16 '12 at 1:33
  • $\begingroup$ No, but it is easy to check whether $4a^{3} + 27b^{2}$ has the required form, and if it does, then the roots are easy to find. I am suggesting a way to "bypass" Cardano's formulae, I am not saying that this method makes Cardano's formulae any easier to use. For example, with $x^{3}- 7x-6$, we find that $4a^{3} + 27b^{2} = -400 = -20^{2}$, so we know that the roots are all integers just from the coefficients of the cubic. Then the roots are three factors of $6$ whose sum iz zero. $\endgroup$ – Geoff Robinson May 16 '12 at 6:18

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