I've been working my way through Cardano's formula for the cubic equation and I'm getting stuck at the end.

I'm clear on producing the depressed cubic, $$y^3 + Ay + B,$$ and from there, I'm clear on substituting $y = s + t$, on getting the system of equations, \begin{align} s^3 + t^3 &= -B\\ 3st &= A \end{align}

Clear on producing and solving the degree 6 equation, $$t^6 - B t^3 + \frac{A^3}{27} = 0.$$

\begin{align*} t^3 &= \frac{B}{2} \pm \sqrt{\left(\frac{B}{2}\right)^2 - \left(\frac{A}{3}\right)^3}\\ s^3 &= \frac{B}{2} \mp \sqrt{\left(\frac{B}{2}\right)^2 - \left(\frac{A}{3}\right)^3}. \end{align*}

$t^3$ has three cube roots, namely $t$, $t \omega$ and $t \omega^2$. And $s^3$ has has three cube roots, namely $s$, $s \omega$ and $s \omega^2$.

As $y$ is a sum of these cube roots, we have a total of 9 possibilities. The below chart shows the relationship between these 9 roots as dictated by the above equation, $3st = A$.

enter image description here

And this is where I get shaky. I can't convince myself that the true solutions are always the ones of the red path. Is there no ambiguity there? What if $t^3$ and $s^3$ are imaginary? Is there no other piece of information needed to definitively say what the roots are? It seems that every write up of this proof skips over this fact. I know I'm missing something here.

I appreciate any insight you folks can provide.


1 Answer 1


You can't say that it's always the red path. Up to the point when you get the equations with $s^3$ and $t^3$, you have nothing to distinguish the three cube roots of $s^3$, or of $t^3$.

Suppose you were to replace $s$ by $s' = s\omega$ (while keeping $t' = t$), and draw a new diagram with $s'$ and $t'$. Then the new red $s' + t' = s \omega + t$ which is the old blue. But the original equation hasn't changed. So if in the new diagram, the red path is correct, then in the old diagram, the blue path was correct.

To choose the correct path, you have to choose $s$ and $t$ so that $3st = A$ is satisfied. If you choose a wrong path, then $3st$ will be $A\omega$ or $A\omega^2$ instead.

  • $\begingroup$ Thanks Ted!. So you have to do some computation here.That's what it looked like but I thought I was missing some clever trick that would tell me which cube roots to pick. If not, at least you only have to compute and check two of those nine choices. $\endgroup$
    – Bo Rel
    Commented Jun 6, 2017 at 16:43

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