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?