The depressed cubic equation $y^3 +py + q = 0$ can be solved with Vieta's transformation (or Vieta's substitution)
$y = z – \frac{p}{3 \cdot z}.$
This reduces the cubic equation to a quadratic equation (in $z^3$).
Is there any geometric or algebraic motivation for this transformation? I am not asking why this transformations works – this is just an easy calculation. I would rather like to know how to come up with it. Perhaps even how and when Vieta came up with it. I haven't found anything about the history of this transformation, except that it probably wasn't invented by Vieta.
Notice that the Ansatz $y = z + \frac{c}{z}$ for a constant $c$ will eventually lead to $c = -\frac{p}{3}$, but what motivates this Ansatz – except for that it works in the end? Here is what I guess (but this is not convincing yet): Polynomial transformations do not work, so let's try rational transformations. Try to keep the degree low.
I am aware of Galois theory and how it helps to understand the cubic from a highly conceptual point of view, but I would like to avoid Galois theory here.
Any information about the history of this transformation will also be appreciated.
Best Answer
Viète's motivation came from the solution of the following problem:
Find two numbers when their sum and the sum of their cubes are given, i.e., find $x$ and $y$ such that $$x+y=a, \qquad x^3+y^3=d$$ A simple solution is to factor the last expression: $$\begin{align}x^3+y^3&=(x+y)(x^2-xy+y^2) \\ &=(x+y)((x+y)^2-3xy) \\ &=a(a^2-3xy)=d\end{align}$$ We can now find $$xy={a^3-d \over 3a}$$ and the problem is reduced to the much simpler one of solving the equations $x+y=a$ and $xy=b$, which Viète had already considered.
Viète now realized that he had a very nice relation between the sum of two numbers, the product of the two numbers, and the sum of their cubes, given by $$a^3 - 3ba = d$$ Hence, if he had an equation on this form, where $a$ was the unknown, he could solve it by finding $x$ and $y$ such that $a=x+y$ and $b=xy$. This leads to the substitution $$a = x + {b\over x}$$ The equation then becomes $$x^3+{b^3\over x^3}=d$$ which gives $d$ as the sum of the cubes, and which is a quadratic in $x^3$.
Edit: The Analytic Art by Viète is available online. The motivating example is on p. 110, and this solution of the cubic equation is on p. 289. Viète's methods are also explained in Victor Katz: A History of Mathematics. An Introduction.