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,

$$(u-x_1)^{1/3}+ (u-x_2)^{1/3}+ (u-x_3)^{1/3} = {v}^{1/3}\tag1$$

This is given by,

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

$$v = \frac{27p^2q^3}{p^6+9p^3q^2+27q^4}$$

Note that $u$ satisfies the rational Diophantine equation,

$$u^3+pu+q=w^3\tag2$$

Eq. $(2)$, being transformable to an elliptic curve, has infinitely many rational points $u$. However, not all of them will yield rational $v$.

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

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?