Understanding Cardano's method

Question

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?

Answer 1

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 $t^3 + pt + q=0$.

Answer 2

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.

Answer 3

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.