Inspiration behind substitution of main variable in Cardano's solution

Question

Suppose the solutions to a general cubic equation $ax^3+bx^2+cx+d=0$ are to be found. Then according to Cardano's method, First a variable substitution must be carried on to convert the general cubic to depressed cubic. $$ax^3+bx^2+cx+d=0\rightarrow t^3+pt+q=0\ \text{where $t=x-\frac{b}{3a}$}$$ This seems pretty clear but what happens next is not so obvious for me.

Now we take $t=u+v$ to get $u^3+v^3+(3uv+p)(u+v)+q=0$. Assuming $3uv+p=0$, we get the system $$u^3+v^3=-q$$ $$uv=\frac{-p}{3}$$. Using Vieta's formula, the quadratic equation with $u^3$ and $v^3$ as roots is $$x^2+qx-\frac{p^3}{27}=0$$ Now the roots for quadratic can be found by analysing the discriminant.

My doubt is that what was the inspiration behind assuming $3uv+p=0$, except helping to form the quadratic equation.

Please help

THANKS

Answer 1

One source of inspiration may have been looking at the binomial expansion of $(u+v)^3$ and mapping it to the cubic $t^3 + pt +q = 0$.

$$\begin{align*}(u+v)^3 &= u^3 +3 u^2v +3uv^2+v^3 \\ \\ &= u^3+v^3+3uv(u+v) \\ \\ (u+v)^3 - 3uv(u+v)-(u^3+v^3)&= 0\\ \end{align*}$$

Substituting $t =u+v$ yields

$$t^3-3uvt-(u^3+v^3) =0$$

Which forces the equalities $$ p = -3uv$$ $$q= -(u^3+v^3)$$ to complete the mapping of the depressed cubic equation to the equation for the cube of a binomial.

Is this what really happened historically? I don't know. Probably not.

Answer 2

There is a geometric picture behind this. The expression $(u+v)^3$ can be visualized as a cube subdivided into eight rectangular solids of sides $u\times u\times u$, $u\times u\times v$ (permuted three ways), $u\times v\times v$ (permuted three ways), and $v\times v\times v$. As an alternative, however, it can be understood as subdivided into five rectangular solids of sides $u\times u\times u$, $t\times u\times v$ (cyclically permuted three ways), and $v\times v\times v$.

If $p$ and $q$ are negative, the equation $t^3+pt=-q$ can be understood geometrically as saying that if from a cube of side $t$ a rectangular solid of length $t$ and cross-sectional area $-p$ is removed, or three rectangular solids of length $t$ and cross-sectional area $-\frac{p}{3}$ are removed, then the remaining volume is $-q$. If we arrange for the cross-sectional area $-\frac{p}{3}$ to be that of a $u\times v$ rectangle, then the remaining volume will consist of a cube of side $u$ and a cube of side $v$. Hence we have $$ v=-\frac{p}{3u},\quad u^3+v^3=-q, $$ which is quadratic in $u^3$.

This doesn't come completely out of the blue: at the point in time when Tartaglia solved the depressed cubic, there was a 3500 year history of solving quadratics using an analogous 2-dimensional geometric picture. The ancient Mesopotamians, for example, solved, in essence, the problem of finding $t$ given that $t(t+a)=b$. They did this by arranging a square of side $t$, a square of side $\frac{a}{2}$, and rectangles of dimensions $t\times\frac{a}{2}$ and $\frac{a}{2}\times t$ into a square of side $t+\frac{a}{2}$, which must then have area $b+\left(\frac{a}{2}\right)^2$. From $\left(t+\frac{a}{2}\right)^2=b+\left(\frac{a}{2}\right)^2$, one can compute $t+\frac{a}{2}$ and hence $t$. Although the Mesopotamian algorithms were described in words only, accompanied neither by equations nor by geometric diagrams, a careful analysis by Jens Høyrup of the geometric terminology used provides convincing evidence that this was how they understood their procedures. Closer to the time of Tartaglia, al-Khwarizmi solved quadratics by methods based on a similar geometric picture.