The title of this post is the one from the source that inspired me: a draft of Lecture Notes of H.M. Khudaverdian , "Galois theory", Manchester, Autumn 2006 (version 16 XII 2006). See pages 76-77.
My concern for this problem appeared exactly one year ago when I saw the Facebook post of Mahan Gholami. I have published about them here and here, but now we will give a treatment of the general case.
As an irony, I posted the proposed problem on our Facebook Page and received all kinds of responses.
Let's take for example the equation x^3-1=0. Its roots are x_1=1,\;x_{2,3}=\frac{-1\pm \imath \sqrt{3}}{2}. The x_2 and x_3 cannot be rationally expressed in terms of x_1=1. However, if we denote by \xi one of the roots x_{2,3} we have
\xi,\;\xi^2,\;-\xi-\xi^2
as the complete list of roots. In our situation it will be seen that we do not have to choose a preferential root, but the expressions are possible whatever the chosen root. The above situation occurs because the polynomial X^3-1 is not irreducible.
The situation is clarified by the Proposition (not named as such) on page 35 of the work cited at the beginning. Combining the ideas from there we enunciate:
Proposition Let f be irreducible cubic polynomial over \mathbb{Q}.
If (and only if) square root of its discriminant are rational number, then
the roots of equation f(x)=0 are rationally expressible via each other.
Remark If one of the roots is \alpha then, because the degree of expansion
of the decomposition field of the polynomial will be equal to 3, the other
roots have the expression
l+m\cdot \alpha +n \cdot \alpha^2 \tag{0}
with the coefficients l,\;m,\;n\;\in \mathbb{Q}.
We will present here only the METHOD by which, in the given case, the stated expression can be found.
Without losing generality, we assume f=X^3+p\cdot X+q \;\in \mathbb{Q}[X] irrreducible. Its roots x_{1,2,3}\;\in \mathbb{C} satisfy the condition
x_1+x_2+x_3=0. \tag{1}
Let
d=(x_1-x_2)(x_2-x_3)(x_3-x_1) \tag{2}
and D=d^2 its discriminant. Here
D=-p^3-27q^2 .\tag{3}
If we denote by \alpha one of the roots, let it be x_1, we will show that the others are expressed by
x_{2,3}=\frac{-3\alpha^3-p\alpha \pm \sqrt{D}}{6\alpha^2+2p}.\tag{4}
From (1) and (2) we have the equation
(\alpha-x_2)(x_2+x_2+\alpha)(-x_2-\alpha-\alpha)=d
which is written after multiplications
2x_2^3+3\alpha x_2^2-3\alpha^2x_2-d-2\alpha^3=0. \tag{5}
On the other hand, from the second Vieta's formula x_1x_2+x_2x_3+x_3x_1=p we deduce \alpha x_2+\alpha (-x_2-\alpha)+x_2(-x_2-\alpha )=p, so
x_2^2+\alpha x_2 +\alpha ^2+p=0. \tag{6}
Multiplying equation (6) by 2x_2+\alpha we get
2x_2^3+3\alpha x_2^2+(3\alpha^2+2p) x_2+3\alpha^3+p \alpha =0. \tag{7}
Subtracting the equations (7) and (5) results
(6\alpha^2+2p)x_2+\alpha^3+p\alpha +d+2\alpha^3=0
where from
x_2=\frac{-3\alpha^3-p\alpha -d}{6\alpha^2+2p},
and then x_3=-x_2-\alpha=\frac{3\alpha^2+p\alpha+d -6\alpha^3-2p\alpha}{6\alpha^2+2p}=\frac{-3\alpha^3-p\alpha+d}{6\alpha^2+2p} i.e. exactly formula (4).
\blacksquare
REMARK CiP
The fact that formula (4) can be expressed in the form (0) results from the following:
All powers of \alpha have the form (0) :
\alpha^3=-p \cdot \alpha -q,
\alpha ^4=-p \cdot \alpha^2-q \cdot \alpha, an so on...;
more, from \alpha^3+p\alpha+q=0\;\;\Rightarrow\; \alpha(\alpha^2+p)=-q so
\frac {1}{\alpha}=-\frac{1}{q} \cdot \alpha^2-\frac{p}{q}.
Moreover, any rational expression R(\alpha)=\frac{P(\alpha)}{Q(\alpha)} with P,\;Q\in \mathbb{Q}[X] can be reduced first to one of the form \frac{P_1(\alpha)}{Q_1(\alpha)} with degree polynomials P_1 and Q_1 at most 2. Then to the form (0), for example with the method of undetermined coefficients.
Let's try \frac{1}{6\alpha^2+2p} that appears in formula (4).
\frac{1}{6\alpha^2+2p}=l+m \cdot \alpha +n \cdot \alpha^2\;\;\Leftrightarrow
\Leftrightarrow\;\;1=(l+m\alpha+n\alpha^2) \cdot (6\alpha^2+2p)\;\;\Leftrightarrow
\Leftrightarrow \;\;1= 6l\alpha^2+2pl+6m \alpha^3+2mp\alpha+6n \alpha^4+2pn\alpha^2\;\;\Leftrightarrow
\Leftrightarrow\;\;1=6l\alpha^2+2pl+6m(-p\alpha-q)+2mp\alpha+6n(-p\alpha^2-q\alpha)+2pn\alpha^2\;\;\Leftrightarrow
\Leftrightarrow\;\;1=\alpha^2 \cdot(6l-4pn) +\alpha \cdot (-4pm-6qn)+(2pl-6qm).
Let's then impose the requirements
\begin{cases}6l-4pn=0\;,\\-4pm-6qn=0\;,\\2pl-6qm=1 \;.\end{cases}
Solving this system of equations we get
n=\frac{3p}{4p^3+27q^2},\;m=-\frac{9q}{2(4p^3+27q^2)},\;l=\frac{2p^2}{4p^3+27q^2}.
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