Motivation of Vieta's transformation

Question

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.

Answer 1

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.

Answer 2

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.

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[ 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 $\,\sqrt[3]{b} = -\frac{p}{3 z}\,$.

Answer 3

Here is another way to discover Vieta's transformation. Start with the classical Chebyshev polynomial $T_n(X)$, which satisfies $T_n(\cos\theta)=\cos n\theta$. It is well-known that

$$T_n\Bigl(\frac{X+X^{-1}}2\Bigr)=\frac{X^n+X^{-n}}2.$$

[One way to deduce this is to write $z:=e^{i\theta}$, so that $\cos\theta=(z+1/z)/2$ and $\cos n\theta=(z^n+1/z^n)/2$, which means the above identity holds when $X$ takes each of the infinitely many values $e^{i\theta}$ -- so subtracting the right side of the identity from the left yields a rational function whose numerator has infinitely many roots, and hence is the zero polynomial.]

In case $n=3$, this identity says

$$(4X^3-3X)\circ\frac{X+X^{-1}}2 = \frac{X^3+X^{-3}}2,$$

or equivalently

$$ (X^3-3X)\circ (X+X^{-1}) = X^3+X^{-3},$$

which yields Vieta's substitution in case $p=-3$. To obtain similar substitutions for an arbitrary $X^3+pX$, just conjugate the above identity by $rX$ to get

$$ \Bigl(rX\circ (X^3-3X)\circ X/r\Bigr) \circ \Bigl(rX\circ (X+X^{-1})\circ X/r\Bigr) = rX\circ (X^3+X^{-3})\circ X/r,$$

which simplifies to

$$ (r^{-2}X^3-3X) \circ (X+r^2X^{-1}) = r^{-2}X^3+r^4X^{-3}.$$

Now multiply by $r^2$ to get

$$ (X^3 - 3r^2 X) \circ (X+r^2 X^{-1}) = X^3 + r^6 X^{-3}.$$

Writing $p:=-3r^2$, this becomes

$$ (X^3 + pX) \circ \Bigl(X-\frac{p}{3X}\Bigr) = X^3 - \frac{p^3}{27X^3},$$

which is Vieta's identity.

So one can view Vieta's substitution as being a change of variables of the identity $\cos 3\theta=4\cos^3\theta-3\cos\theta$.

I note that one can similarly solve any equation of the form $T_n(X) = q$ by substituting $(X+1/X)/2$ for $X$ to get $X^n + X^{-n} = 2q$, yielding a quadratic polynomial in $X^n$. The unusual feature of the case $n=3$ is that every degree-$3$ polynomial is a linear change of variables of either $X^3$ or $T_3(X)$, so that the fact we can solve $T_n(X)=q$ and $X^n=q$ implies we can solve every degree-$3$ polynomial.

Answer 4

In the notation of $$p=-\dfrac34r^2\tag1$$ the given cubic takes the form of $$4y^3-3r^2y+4q=0,$$ or $$4\,\dfrac{y^3}{r^3}-3\,\dfrac yr + \dfrac{4q}{r^3}=0.\tag2$$ Let $$y=r\cosh t =\dfrac r2\left(e^t+e^{-t}\right),\tag3$$ then from $(2)-(3)$ should $$\cosh 3t + \dfrac{4q}{r^3}=0,\quad \dfrac{r^3}8\left(e^{3t}+e^{-3t}\right)+q=0.\tag4$$ Assuming $$z=\dfrac r2\,e^t,$$ easily to get $$y=z+\dfrac{r^2}{4z}=z-\dfrac p{3z},$$ $$z^3-\dfrac{p^3}{27z^3}+q = 0.$$ This makes clear the Vieta's substitution.