TITLE: An elliptic curve for Ramanujan-type cubic identities?
QUESTION [8 upvotes]: Given the roots $x_i$ of the depressed cubic,
$$x^3+px+q=0$$
with rational coefficients. It can be shown that, in general, one can find rational $u,v$ such that,
$$(u-x_1)^{1/3}+ (u-x_2)^{1/3}+ (u-x_3)^{1/3}  = {v}^{1/3}\tag1$$ 

Solution 1:

$$u = \frac{-p^6q+27q^5}{p(p^6+9p^3q^2+27q^4)}$$ 
$$v = \frac{27p^2q^3}{p^6+9p^3q^2+27q^4}$$

Solution 2:

$$u=\frac{p^{12} - 135 p^6 q^4 - 729 p^3 q^6 - 729 q^8}{9 p q (p^3 + 6 q^2) (p^6 + 9 p^3 q^2 + 27 q^4)}$$
$$v =\frac{3 (p^8 + 12 p^5 q^2 + 36 p^2 q^4)}{q (p^6 + 9 p^3 q^2 + 27 q^4)}$$
and so on, with the $u,v$ becoming increasingly long. (One $u$ was a rational $30$-deg poly in $p$.)
Note: The $u$ satisfy the rational Diophantine equation,
$$u^3+pu+q=w^3\tag2$$
Eq. $(2)$ can be transformed to an elliptic curve. However, not all of its rational $u$ will yield rational $v$.

Question: How do we show that the two $u,v$ above, in fact, are just the first pairs of infinitely many rational solutions?

REPLY [6 votes]: The following statement can be viewed as an addendum to Zieve's answer.
Proposition. Let $p$ and $q$ be rational numbers such that
$$ p \ne 0,\quad q \ne 0 , \quad p^3+6q^2 \ne  0,\quad p^3+9q^2 \ne 0 . $$
Then the curve $E : y^2=x^3+27p^3+\frac{729}4q^2 $ has infinitely many rational points. 
Proof. Let us consider two cases.
Case 1: $27p^3+\frac{729}4q^2 = 0$; then one can obtain infinitely many rational points on $E : y^2=x^3 $ via a rational parametrisation.
Case 2: $27p^3+\frac{729}4q^2 \ne 0$; then $E$ is a nonsingular cubic with a point $P:= (-3p : \frac{27}{2} q : 1) $, that is, an elliptic curve (we use homogeneous coordinates here). 
It is clear that $P\ne O$, where $O:=(0:1:0)$ is the point at infinity. Futhermore, the order of $P$ can be $2$, $3$ or $6$ (see [1, p.134, Theorem 5.3] ). Using standart computer software one can obtain
$$ 2P=(  pq(p^3 + 6q^2)  : -(p^6 + 9p^3q^2 + \frac{27}2q^4)  : q^3  ),$$
$$ 3P=( -3p(p^3 + 9q^2) (p^9 - 81p^3q^4 - 243q^6)  : \frac{81}2 q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) :  p^3(p^3+9q^2)^3 ),$$
$$ 6P=( \frac19 qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)(p^9-...)(p^{27}+...)   : -\frac1{27}(p^6+...)(p^{12}+...)(p^{36}+...)  :  q^3p^3(p^3 + 6q^2)^3(p^3 + 9q^2)^3(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)^3 ).$$
Let us prove that in fact the point $P$ is of infinite order in the group $E(\mathbb Q)$.

Suppose $2P=O$; this means that
$$    (  pq(p^3 + 6q^2)  : -(p^6 + 9p^3q^2 + \frac{27}2q^4)  : q^3  ) \sim (0:1:0)  $$
or equivalently,
$$ \left\{ \begin{aligned} &q=0 \\  &p^6 + 9p^3q^2 + \frac{27}2q^4 \in \mathbb Q^{\times}\\ \end{aligned} \right.             $$
But $q \ne 0$ by hypothesis, hence the system above has no solutions.
Suppose $3P=O$; reasoning as above, we get 
$$ \left\{ \begin{aligned} &p(p^3+9q^2)=0 \\  
&q (p^3 + 6q^2) (p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6) \in \mathbb Q^{\times}\\ \end{aligned} \right.             $$
The constraints $p \ne 0$ and $p^3+9q^2 \ne 0$ ensure that this system has no solutions.
Suppose $6P=O$; similary we have
$$ \left\{ \begin{aligned} &qp(p^3 + 6q^2)(p^3 + 9q^2)(p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6)=0 \\  &(p^6+...)(p^{12}+...)(p^{36}+...)    \in \mathbb Q^{\times} \\ \end{aligned} \right.             $$
To complete the proof of the proposition it suffices to show that the equation 
$$  p^9 + 9p^6q^2 + 27p^3q^4 + 81q^6  =0                 $$
has no solutions over $\mathbb Q$.  

Assume that there exists such a rational solution $(p,q)$. 
Then $(\tilde p,\tilde q):=(p^3+3q^2 , -3q^2)$ is a rational solution of 
$\tilde p^3=2 \tilde q^3$, which is a contradiction.

Aside: We omitted the coordinates of $6P$ for reasons of space. The following Magma code can be useful.
F<p,q> := FunctionField(Rationals(),2);
E:=EllipticCurve([0, 27*p^3+729/4*q^2]); 
P:=E![-3*p,27/2*q,1];
"the coordinates of 6P:"; 
6*P; a:=(6*P)[1]; b:=(6*P)[2];
"the numerator of the first coordinate of 6P:";
Factorisation(Numerator(a)); 
"the denominator of the first coordinate of 6P:";
Factorisation(Denominator(a));
"the numerator of the second coordinate of 6P:";
Factorisation(Numerator(b)); 
"the denominator of the second coordinate of 6P:";
Factorisation(Denominator(b));

References:
[1] Knapp, A. W. (1992). Elliptic curves (Vol. 40). Princeton University Press.