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Take any cubic real polynomial $f$ with roots $p,q,r$.

Let $x = p^2 q + q^2 r + r^2 p$.

Let $y = p q^2 + q r^2 + r p^2$.

Let $z = p + q + r$.

Then $x+y$ and $xy$ are both symmetric in $p,q,r$ and hence can be expressed in terms of the coefficients of $f$. This implies that we can solve for $x,y$ via the appropriate quadratic equation.

But after that, is it possible to algebraically recover the roots $p,q,r$ from $x,y,z$? I think that the three equations uniquely determine them, but I do not see an easy or motivated way to continue. Is this approach viable or is it doomed to failure, and why?

This question was inspired by this post, whose author suggested such an approach but we could not see how to complete it to obtain the roots.

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user21820
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  • By the way, I'm aware of both Cardano's and Lagrange's solution to the cubic, and the Galois theory behind solvability of polynomials. – user21820 Mar 05 '17 at 05:02
  • So what is the goal? Are you trying for an alternate cubic formula? – quasi Mar 05 '17 at 05:32
  • @quasi: The goal in the linked post was to give an intuitive explanation of the unsolvability of the quintic. It actually fails because it does not show that such algebraic expressions are necessary nor sufficient to obtain the roots of a polynomial. Hence my goal here is to understand whether such an approach can be patched, which would naturally give an alternate solution to the cubic. Ultimately the formula is going to be the same, but the route may be different and illuminate something on the way. – user21820 Mar 05 '17 at 05:39
  • I'm not sure I follow, but I can help a little with some of the calculations. I'll post what I have as an answer in a little while (it would be only a few minutes if I were a more competent typist). – quasi Mar 05 '17 at 05:53
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    IMHO a key Galois theoretic aspect in Cardano's solution is to find elements of the splitting field that have no symmetries but whose cubes are invariant under the cyclic shifts (IOW the cubes are in the fixed field of $A_3$). I try to explain that here - probably nothing new to you, just the standard way of finding enough linear combinations of the zeros in a certain root tower extension. – Jyrki Lahtonen Mar 05 '17 at 06:37
  • If a quantity $z$ not stable under the 3-cycle $\sigma=(123)$ of roots has the property $\sigma(z^3)=z^3$, then $\sigma(z)/z$ is a primitive third root of unity. I think we need a counterpart of that to make progress. Of course, you can work with angle triplication trig formulas also. Anyway, the point I'm getting at is that if $K$ is the field of coefficients, $L$ is the splitting field (so generically $[L:K]=6$), and $F=\operatorname{Inv}(A_3)$ is the quadratic intermediate field, $[F:K]=2$, $F=K(x)=K(y)$, then we still need to exhibit a number $w$ such that $L=F(w)$. – Jyrki Lahtonen Mar 05 '17 at 06:46
  • Cardano's method (implicitly) uses $w=p+\omega q+\omega^2 r$ or the equivalents, when $\sigma(w)=\omega w$. We need an analogue of that - a way to "break the symmetry between $p,q,r$ to isolate and solve them". – Jyrki Lahtonen Mar 05 '17 at 06:52
  • @JyrkiLahtonen: Thanks for that! I always thought that was Lagrange's solution and not Cardano's, but I remembered wrong; apparently the one I had learnt was Harriot's, and then later I learnt Lagrange's. – user21820 Mar 05 '17 at 10:41
  • I'm no historian, so don't quote me on calling that exactly Cardano's method :-). My answer to that other question was an adaptation of what I learned from Jacobson's Basic Algebra I. Anyway, all the methods I know of for solving eventually turn the equation into a linear system on the appropriate DFT of the unknown roots: the method for solving quadratic first finds $x_1+x_2$ and (the square of) $x_1-x_2$. Ferrari's method for quartics has an explanation in terms of $x_1+ix_2-x_3-ix_4$ et cetera. – Jyrki Lahtonen Mar 05 '17 at 10:49
  • @JyrkiLahtonen: Yup I see that structure in the quartic solution as well. My question is just a curiosity since Harriot's solution is kind of different, though completely ad-hoc, as if there was no deep reason for it to work. – user21820 Mar 05 '17 at 10:51

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Suppose the cubic has the form

$$f(x) = t^3 + bt - c$$

shifted in advance so that the coefficient of $t^2$ is $0$.

Letting $p,q,r$ denote the roots, and let

\begin{align*} x &= p^2q + q^2r + r^2p\\[6pt] y &= pq^2 + qr^2 + rp^2\\[6pt] z &= p + q + r \end{align*}

Since the coefficient of $t^2$ is $0$, it follows that $z=0$.

Let $u = x + y$ and let $v = xy$.

Using Maple, I derived the following relations:

\begin{align*} &u = -3c\\[3pt] &v = b^3+9c^2\\[14pt] &x^2+(3c)x+(b^3+9c^2) = 0\\[3pt] &y^2+(3c)y+(b^3+9c^2) = 0\\[14pt] &c = -\frac{x+y}{3}\\[3pt] &b^3 = -(x^2 + xy + y^2)\\ \end{align*}

But recovering $p,q,r$ from $x,y$ is not really viable$\,-\,$you get a $9$-th degree equation (although it's effectively cubic):

$$27p^9+27(x+y)p^6-9(2x^2+xy+2y^2)p^3+(x+y)^3 = 0$$

and the same equation for the other two roots.

quasi
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  • That was essentially the issue I had with the linked post. I know how to find $x,y$ from the coefficients, but don't see a viable way to get from there to the roots without having to start all over again with a new cubic. – user21820 Mar 05 '17 at 06:23
  • I don't really follow the point of the linked post, although I just looked at it briefly. It seems as if the OP is claiming the unsolvability of the quintic can be explained by somehow side-stepping Galois Theory. I'm not sure I really buy that. I suspect if you look closely at the argument (which admittedly I haven't done), the Galois Theory is there in camouflage, and moreover, the camouflage makes it both less understandable and more "mystical" (i.e., unnecessarily mysterious). – quasi Mar 05 '17 at 06:29
  • Yes I don't buy the possibility of side-stepping Galois theory, since I grasp the real theory (or at least I once fully did). I felt the linked post would be useful (for an intuitive hand-waving explanation) iff my current question could be answered affirmatively, but I'm not very hopeful about that. – user21820 Mar 05 '17 at 06:32
  • My enquiry is also partly because the linked post got 18 upvotes and no downvotes so somehow 18 people thought it was useful! In the 3 years since it was posted, I'm the first to point out the rather huge gap in the 'explanation', so either 18 people were not paying attention or I am the one missing something obvious. – user21820 Mar 05 '17 at 06:34
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    Mystical explanations tend to "wow" people. There's also the "emperor's clothes" phenomenon. Also, it's hard to disprove the OP's claims, hence the lack of downvotes. – quasi Mar 05 '17 at 06:35
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    Hahaha! So I'm the little child? =D – user21820 Mar 05 '17 at 06:36
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    It's good that someone is! – quasi Mar 05 '17 at 06:38