# Which cube roots do we take in the cubic formula?

I've been studying Cardano's method lately, and there's one step that I haven't been able to convince myself is correct.

The method relies two times on the following fact:

Lemma: For $a, b\in \Bbb C$, there exists a unique pair $u, v$ with $u + v=a$ and $uv=b$.

We start with a depressed monic cubic having some root $x$:

$$x^3+ax+b=0$$

Obviously there exists $u, v$ with $u+v=x$, and when we expand with the binomial formula and gather terms, we can obtain:

$$u^3+v^3+b+(u+v)(3uv+a)=0$$

By our lemma, there exists a unique pair such that on top of $u+v=x$, we also have $uv=-a/3$. In other words, the following system (1) holds:

\begin{align}u+v&=x\\ uv&=-a/3\end{align}\tag{1}

For that pair, we get (2):

\begin{align}u^3+v^3&=-b\\ u^3v^3&=-a^3/27\end{align}\tag{2}

By our lemma again, we can get expressions $u^3$ and $v^3$. It is then claimed that $u$ and $v$ are gotten by taking the cube roots of these expressions. That's the step that seems unclear to me, especially considering each one has $3$ cube roots anyway.

Why is it that if $u, v$ satisfy system (2), they must satisfy system (1)? Or, do the cube roots in the cubic formula refer to a specific choice of cube root? If so, how do you choose the correct one?

You have three choices for $u$ and independently three choices for $v$ to satisfy $u^3+v^3=-b$ and $u^3v^3=-a^3/27$. But for each choice of $u$, you have no more choice for $v$ to satisfy $uv=-a/3$. So ultimately you have only three choices,whic is fine as there are three roots.
• But there are choices of $u$ and $v$ which might not give a solution to the original cubic, right? And are all the roots of the original cubic to be found among the choices of $u$ and $v$? – Jack M Dec 13 '13 at 11:15
• Yes, if you run over all $9=3\cdot 3$ choices, you get spurious solutions because $u^3v^3=-a^3/27$ does not imply $uv=-a/3$. By construction, for any number $x$ that is a solution (we know there are in general three in $\mathbb C$), we find suitable $u,v$; and in fact we see that if $u+v$ is one root, then the others must be $\zeta u+\zeta^2v$ and $\zeta^2u+\zeta v$ with $\zeta=-\frac12+\frac{i\sqrt 3}2$: By what you wrote the pair $\{u^3,v^3\}$ is unique so that only the choices we have in computing $u$ from $u^3$ (and then necessarily $v=-\frac a{3u}$) can lead to the different solutions. – Hagen von Eitzen Dec 13 '13 at 12:55
• You write $v=-a/3u$. But (this has been bothering me for a while) is the possibility that $u=0$ not a problem? I realize that implies $a=0$ and thus means the cubic is pretty trivial anyway, but the cubic formula is supposed to be general. – Jack M Dec 13 '13 at 21:39
• @HagenvonEitzen, your comment hits on exactly what I had been stuck on. $u$ and $v$ are not independent but in fact determined by one another and wlog, you can let $u$ act as the independent variable. However, how is it clear that $u + v$ is 'main' root, and not $\zeta u + v$ or $\zeta^2 u + v$? The overall proof of Cardano's formula lends itself to 9 possible answers, 3 groups of 3 roots. But it just isn't clear to me, without having to do the actual computation, which group of three to choose from. – Bo Rel Jun 5 '17 at 4:27