I was recently reading about Cardano's method for solving the cubic of the form $t^3 + pt + q = 0 $.

So, you substitute $t$ with two linear variables $t = u+v$. You get the equation $u^3 + v^3 + (3uv +p)(u+v) +q = 0$. Everything is okay thus far. However, then Allan Clark, whose book I was reading, states:

Since we have substituted two variables, $u$ and $v$, in place of the one variable $t$, we are free to require $3uv + p = 0$.

He writes this as if it is the most obvious thing in the world, but I cannot understand why this is true.

So, how do you justify that step?


3 Answers 3


The stroke of genius is putting $t=u+v$.

We have $(u+v)^3= 3uv(u+v)+(u^3+v^3)$, or $$(u+v)^3-3uv(u+v)-(u^3+v^3)=0$$

Compare this with $$t^3 + pt + q=0$$ If $3uv=-p$ and $u^3+v^3=-q$, then $t=u+v$ solves $t^3 + pt + q=0$.

  • $\begingroup$ Woah! Seeing that comparison, it finally makes sense! Thank you very much! $\endgroup$
    – Avatrin
    Jun 10, 2015 at 20:14

The basic idea is that by adding a degree of freedom to the equation, you can use that extra degree to impose a secondary constraint.

Specifically, rewrite $v = t-u$. Then if you require $3uv+p = 0$, that is equivalent to

$$ 3u(t-u)+p = 0 $$ $$ 3u^2-3tu-p = 0 $$

and you can solve for $u$ in terms of $t$ and $p$. So it is feasible: the situation is not overdetermined.

  • $\begingroup$ Why can you do that? What is the underlying logic? Clark uses it to remove v from the equation.. I should add.. My understanding of substitution is abysmal. $\endgroup$
    – Avatrin
    Jun 10, 2015 at 18:49
  • $\begingroup$ I've explained why you can do it. As to why you do it—well, that's the role of inspiration in mathematics. You do it because it works. A lot of mathematics is written so that you only see what works. I'm sure there was a lot of stuff that didn't work on the cubic that you never get to see. $\endgroup$
    – Brian Tung
    Jun 10, 2015 at 19:03

You can impose a condition on $u$ and $v$, since only one condition has been specified thus far. ($t=u+v$)

This would still uniquely determine $u$ and $v$. I am guessing that the authors chose this specific condition $3uv+q=0$ because it makes the later calculations easier.

However, after the two conditions have been specified, another condition cannot be, since it may then be the case that the three equations in two unknowns have no solutions.

So the number of conditions that you can specify is equal to the number of variables you are dealing with.


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