See my previous question for a refresher on how Cardano's method for the cubic works. The gist is that where $x$ is any root of the cubic, we know that $x = u+v$, where we have explicit formulae for $u^3$ and $v^3$. So now we want to take cube roots of $u^3$ and $v^3$, and add them together to form $x$.

We can't just take any cube roots. However, what we can do is take an arbitrary cube root of $u^3$, and then since we know:


we can conclude that $v=\frac{-n/3}{u}$. But how do we know $u\neq0$?

I get that we can suppose, since the expressions involving $u$ and $v$ are symmetric, that $u$ is non zero as long as we can guarantee that at least one of $u$ or $v$ is non-zero. I can also see that if both $u$ and $v$ are zero, then the cubic is just $x^3=0$, which is trivial. But still, the cubic formula is supposed to be general. Not only that, but just plugging in zero coefficients to the cubic formula does yield a correct solution, so clearly this division by zero isn't actually a problem. Why is that?

  • $\begingroup$ Just to clarify, you're saying the formula should be general, but you're concerned that it doesn't hold for $u,v$ both equal to zero? $\endgroup$ – Eric Auld Dec 18 '13 at 13:23
  • $\begingroup$ @EricAuld Yes. And therefore doesn't hold when the non-leading coefficients are both $0$. $\endgroup$ – Jack M Dec 18 '13 at 14:24
  • $\begingroup$ I will say that it would not be the first time that an "official" source has neglected to include such an assumption. (Like they perhaps should have said "Suppose $a\neq 0$..." at the beginning.) $\endgroup$ – Eric Auld Dec 18 '13 at 15:22

Since, by construction, $u^3v^3 = -a^3/27$, the only way $u$ (or $v$) can be zero is if $a=0$. However, if $a=0$ then the initial cubic equation was just $x^3+b=0$, and then solving it is just a matter of taking a single cube root -- no need for Cardano in the first place.

  • $\begingroup$ Of course - but the cubic formula is supposed to be completely general, just like the quadratic formula. $\endgroup$ – Jack M Dec 18 '13 at 18:59
  • $\begingroup$ @JackM: Who's doing that supposing? $\endgroup$ – hmakholm left over Monica Dec 18 '13 at 19:01
  • $\begingroup$ That's a good point. $\endgroup$ – Jack M Dec 18 '13 at 21:54

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