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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.

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  • $\begingroup$ You may find this helpful: mathforum.org/kb/message.jspa?messageID=306173 $\endgroup$ – Gerry Myerson Nov 9 '15 at 8:41
  • $\begingroup$ Well, this is the connection to Cardano's resp. Tartaglia's method. But these are even less motivated, in my opinion. In texts the story is usually told as: Let's try $y=u+v$ and see what happens. See, it works! But this is no motivation (for me). Ideally, I would like to see an argument of the form: The cubic function has to be composed with $z + \frac{c}{z}$ because ..." $\endgroup$ – Martin Brandenburg Nov 9 '15 at 9:29
  • $\begingroup$ OK, then, try this: maa.org/press/periodicals/convergence/… "I want to think along with Cardano, to understand his solution as he did, and to bring my students to an understanding of Cardano’s world of mathematics—his ways of thinking and the mathematical tools at his disposal. The elucidation of these points is my goal in this article." $\endgroup$ – Gerry Myerson Nov 10 '15 at 1:33
  • $\begingroup$ Have you had a look at the maa link? $\endgroup$ – Gerry Myerson Nov 11 '15 at 5:23
  • $\begingroup$ Yes. It is quite interesting that Cardano's proof uses solid geometry as a motivation. I was hoping for something more analytic. Maybe this is too naive. $\endgroup$ – Martin Brandenburg Nov 11 '15 at 9:28

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