Expressing the roots of a cubic as polynomials in one root

Question

All roots of $8x^3-6x+1$ are real. (*)

The discriminant of $8x^3-6x+1$ is $5184=72^2$ and so the splitting field of $8x^3-6x+1$ has degree $3$.

Therefore, all three roots can be expressed as polynomials in any one given root.

Indeed, if $a$ is a root, then the others are $2a^2-1$ and $-2a^2-a+1$.

This can be easily checked. But how can we find these expressions in the first place?

I've tried this: let $b,c$ be the other roots. Then from Vieta's formulas we get $$b+c=-a, \qquad bc=-\dfrac{1}{8a}$$

The corresponding quadratic has discriminant $3-3a^2$ but it is not at all clear that this is the square of a polynomial in $a$. (It is $(4 a^2+a-2 )^2=(b-c)^2$, as it should be.) I'm stuck here.

More generally, is there an algorithm that given a cubic with integer coefficients and having real splitting field of degree $3$, expresses all three roots as polynomials in any one given root?

(*) From the triple-angle formula $\sin (3\theta) = - 4\sin^3\theta + 3\sin\theta$ when $\sin (3\theta) = 1/2$, these roots are $\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$, but perhaps this is immaterial here.

Undetermined coefficients? The polynomials we need are at most degree 2. — hardmath, May 01 '16 at 20:34
@hardmath, right, but I haven't tried undetermined coefficients because we'll get nonlinear equations for the coefficients... — lhf, May 01 '16 at 20:41
The roots are $a,g(a),g(g(a))$ for $g(x)=2x^2-1$. So the question is how to find $g$, both for the particular equation I gave and also in general. Note that $g(g(g(a)))=a$. — lhf, May 02 '16 at 01:28
http://math.stackexchange.com/questions/794942/express-one-root-of-depressed-cubic-equation-via-another-and-square-root-of-disc seems relevant. — lhf, May 02 '16 at 01:52
This question is similar to: Express one root of depressed cubic equation via another and square root of discriminant. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Bryan, Dec 10 '24 at 13:20
@Bryan, I've already mentioned that question in my answer, but it does not have all details; my answer does. Thanks. — lhf, Dec 10 '24 at 13:47

score 12 · Accepted Answer · edited Apr 13 '17 at 12:21

Based on the answers in this question, I'm able to finish the general case.

Let $x^3+px+q$ be a cubic with rational coefficients and having real splitting field of degree $3$. Let its roots be $a,b,c$. Then $$b+c=-a, \qquad bc=-\dfrac{q}{a}$$ and $b,c$ are the roots of the quadratic $$ h(x)=(x-b)(x-c)=x^2+ax-\dfrac{q}{a} $$ The discriminant of $h$ is $b-c$. Now comes the nice idea from those answers: $$ d=(a-b)(a-c)(b-c)=h(a)(b-c) $$ where $d^2=-4p^3-27q^2$ is the discriminant of the original cubic. The hypothesis on the cubic implies that $d$ is rational. Therefore, $$ b-c = \dfrac{d}{h(a)} $$ To write this as a polynomial in $a$, we solve the linear system on $A,B,C$ implied by $$ d=(Aa^2+Ba+C)h(a)=(Aa^2+Ba+C)(2a^2-\dfrac{q}{a}) $$ or $$ ad=(Aa^2+Ba+C)(2a^3-q)=(Aa^2+Ba+C)(-2pa-3q) $$ The solution is $$ A=-\dfrac{6p}{d}, \quad B=\dfrac{9q}{d}, \quad C=-\dfrac{4p^2}{d} $$ which gives $$ b,c = -\dfrac{a}{2} \pm \dfrac{1}{2}(Aa^2+Ba+C) $$

Nice! Now I need to figure out a way of turning this into an exercise for my course :-) — Jyrki Lahtonen, Mar 10 '18 at 07:41

Expressing the roots of a cubic as polynomials in one root

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