TITLE: Motivation of Vieta's transformation
QUESTION [29 upvotes]: 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.

REPLY [2 votes]: Vieta's own motivation was nicely covered in @PerManne's answer. The following is another take on a plausible algebraic insight, which uses nothing that wouldn't have been familiar to Vieta.
Suppose that a root of the depressed cubic $\,y^3+py+q=0\,$ is in the form $\,y=\sqrt[3]{a}+\sqrt[3]{b}\,$, which would not have been an unreasonable assumption in those times [1].
Then, with $\,u=a+b, v=ab\,$, it follows that:
$$
\begin{align}
y^3 \;&=\; \left(\sqrt[3]{a}+\sqrt[3]{b}\right)^3 \;=\; a + b + 3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})
 \;=\; u +3 \sqrt[3]{v}\,y
\\ & \;\;\;\;\;\;\;\;\iff \;\;\;\;\;\; y^3 - 3 \sqrt[3]{v}\,y - u = 0 \tag{1}
\end{align}
$$
For $\,(1)\,$ to match the original equation, thus ensuring $\,y\,$ is a root, it is sufficient that:
$$
\begin{align}
\begin{cases}
-u &= q
\\ -3 \sqrt[3]{v} &= p
\end{cases}
\;\;\;\;\implies\;\;\;\;
\begin{cases}
u &= -q
\\ v &= -\dfrac{p^3}{27} \tag{2}
\end{cases}
\end{align}
$$
It follows that $\,a, b\,$ are the roots of $\,t^2+qt-\dfrac{p^3}{27}=0\,$, which is the same quadratic derived from Vieta's (and others') methods.
Moreover, letting $\,z=\sqrt[3]{a}\,$, it follows from $\,(2)\,$ that $\,\sqrt[3]{b}=-\dfrac{p}{3z}\,$, so $\,y=\sqrt[3]{a}+\sqrt[3]{b}=z-\dfrac{p}{3z}\,$ not only justifies Vieta's form, but also makes the meaning and symmetry more explicit.
[1]  Quoting Friedrich Katscher from How Tartaglia Solved the Cubic Equation - Tartaglia's Solution:

The quadratic equation has a solution in the form of a binomio $a+\sqrt{b}$ or of a residuo $a−\sqrt{b}$. This suggested that cubic equations might also have a solution in the form of a binomio or of a residuo, however, with cube roots instead of quadratic ones. Probably Tartaglia tried the different possibilities, and thus found that the residuo $\sqrt[3]{a}−\sqrt[3]{b}$ and the binomio $\sqrt[3]{a}+\sqrt[3]{b}$ led to the solutions he sought.


[ EDIT ]   One other less obvious motivation for Vieta's $\,z\,$ substitution in preference to the $\,\sqrt[3]{a}+\sqrt[3]{b}\,$ form used in Tartaglia and Cardano's methods is that it avoids the ambiguity in the choice of cube roots when the roots of the quadratic are (non real) complex. Once one of the cube roots is chosen as $\,\sqrt[3]{a}=z\,$, the other one gets necessarily chosen as $\,\sqrt[3]{b} = -\frac{p}{3 z}\,$.