I have a general question about the cubic equation:
Let $a,b,c \in \mathbb{C}$ with $a \neq 0$. The general cubic equation is $t^3+at^{2}+bt+c = 0$. To get Cardano's Formula, we first transform the equation so that $a= 0$ (i.e. $t^3+bt+c = 0$). Then we reduce this to a quadratic that we can solve. What is the motivation behind the transformations that reduce $t^3+at^2+bt+c = 0$ to $t^3+bt+c = 0$?
In particular, if we let $y = t+ \frac{a}{3}$, then $t = y-\frac{a}{3}$ which cancels the $at^2$ term. This is called a Tschirnhaus transformation. We are left with an equation of the form $$y^3+py+q = 0$$
But how do we know what transformations to use to get Cardano's Formula, where $$p = \frac{a^2-2a^3+3b}{3}\quad\mathrm{and}\quad q = \frac{2a^3-9ab+27c}{27}\quad ?$$ Finally if we let $y = \sqrt[3]{u}+ \sqrt[3]{v}$ we eventually get a quadratic which leads to Cardano's Formula.
Best Answer
I don't think it's really all that mysterious. The motivation behind using the Tschirnhaus Transformation seems clear; it's better to have fewer terms and fewer coefficients when solving an equation. It next probably occurred to Cardano to try $y = u + v$ and see what happens. Rewrite the equation as $$u^3 + v^3 + 3uv(u + v) + p(u + v) + q = 0$$ You have a little leeway in choosing $u$ and $v$, since you only need them to sum to a root of the equation. So you might want to impose a second condition on $u$ and $v$ to simplify things (and two equations in the two variables $u$ and $v$ should specify their values). To reduce the above to a quadratic equation, you can stipulate that $uv = -{p \over 3}$. Namely you have $$u^3 + v^3 + q = 0$$ which is equivalent to $$u^3 -{p^3 \over 27}{1 \over u^3} + q = 0$$ Or just $$u^6 + qu^3 - {p^3 \over 27} = 0$$ This is a quadratic equation in $u^3$, which you can solve. $u$ can be any of the three cube roots of this, and $v$ is determined by the equation $3uv = -p$. Thus you have three possible values of $u + v$ which will be the three roots of your equation.
In Cardano's time they didn't really know much about complex numbers (i.e. that every number has $n$ complex $n$th roots) but he was smart enough to basically understand what was going on.