Let $x_1$,$x_2$,$x_3$ be the roots of the cubic $x^3+px+q$ over $\mathbb Q$, the idea is that rational solutions $(u,v)$ of the equation
$$(u-x_1)^{1/3}+ (u-x_2)^{1/3}+ (u-x_3)^{1/3} = {v}^{1/3} \quad (1)$$
actually lie on a certain affine plane algebraic curve, which can be constructed as follows. First, consider the variety $V$ in the afiine space $\mathbb A^8\langle u,v,x_1,x_2,x_3,y_1,y_2,y_3\rangle$ over the function field $\mathbb Q(p,q)$, $$V: \left\{ \begin{aligned} &x_1+x_2+x_3=0 , \\ &x_1 x_2 + x_2 x_3+x_1 x_3 =p,\\ &x_1 x_2 x_3 =-q, \\ &y_1^3= u-x_1,\\ &y_2^3= u-x_2,\\ &y_3^3= u-x_3,\\ & (y_1 +y_2 +y_3 )^3 = v .\\ \end{aligned} \right. $$
Next, define our curve $C$ as the image of $V$ under the map $$ f : V \to \mathbb A^2 \langle x,y \rangle,$$ $$(u,v,x_1,x_2,x_3,y_1,y_2,y_3) \mapsto (u,v).$$ Finally, get a polynomial defining $C$ with the aid of Magma Calculator:
> F<p,q>:=FunctionField(Rationals(),2);
> A8<x1,x2,x3,y1,y2,y3,u,v>:=AffineSpace(F,8);
> V:=Scheme(A8,[x1+x2+x3, x1*x2+x2*x3+x3*x1-p, x1*x2*x3+q,
> u-x1-y1^3, u-x2-y2^3, u-x3-y3^3, (y1+y2+y3)^3-v]);
> A2<x,y>:=AffineSpace(F,2);
> f:=map<V->A2|[u,v]>;
> C:=Image(f); C;
Scheme over Multivariate rational function field of rank 2 over Rational Field
defined by
x^7*y^2 - 1/27*x^6*y^3 + 1/9*x^5*y^4 + 11/9*p*x^5*y^2 - q*x^5*y - 1/243*x^4*y^5
- 13/27*p*x^4*y^3 + 11/9*q*x^4*y^2 + 2/9*p^2*x^4*y + 1/243*x^3*y^6 +
46/243*p*x^3*y^4 - 16/27*q*x^3*y^3 + 7/27*p^2*x^3*y^2 - 5/9*p*q*x^3*y -
1/27*p^3*x^3 - 1/6561*x^2*y^7 - 10/729*p*x^2*y^5 - 8/243*q*x^2*y^4 -
73/243*p^2*x^2*y^3 + 26/27*p*q*x^2*y^2 + (1/9*p^3 - 7/9*q^2)*x^2*y -
1/9*p^2*q*x^2 + 1/19683*x*y^8 + 35/6561*p*x*y^6 - 7/729*q*x*y^5 +
29/243*p^2*x*y^4 - 155/243*p*q*x*y^3 + (-1/9*p^3 + 19/27*q^2)*x*y^2 +
2/9*p^2*q*x*y - 1/9*p*q^2*x - 1/531441*y^9 + 1/6561*p*y^7 + 53/6561*q*y^6 -
1/243*p^2*y^5 + 20/243*p*q*y^4 + (1/27*p^3 - 28/243*q^2)*y^3 - 1/9*p^2*q*y^2
+ 1/9*p*q^2*y - 1/27*q^3
Note. The bijection between rational solutions $(u,v)$ of equation $(1)$ and rational points $(x,y)$ on the curve $C$ is straightforward: $(u,v) \longleftrightarrow (x,y)$.
Futher, by using formulas in this answer, one can build a rational map (which will be denoted by $\varphi$) from the elliptic curve $E: e^2=d^3+27p^3+\frac{729}4q^2$ to the curve $C$:
> A<d,e>:=AffineSpace(F,2);
> E := Curve(A,[-e^2+d^3+27*p^3+(729/4)*q^2]);
> u:=(1/9*d^6*e - 27/2*q*d^6 - p*d^5*e - 243/2*p*q*d^5 + (-12*p^3 -
> 81*q^2)*d^3*e + (-1458*p^3*q - 19683/2*q^3)*d^3 + (216*p^4 +
> 1458*p*q^2)*d^2*e + (-2592*p^6 - 34992*p^3*q^2 - 118098*q^4)*e)/(d^7
> + 9*p*d^6 + 81*p^2*d^5 + (216*p^3 + 1458*q^2)*d^4 + (972*p^4 +
> 6561*p*q^2)*d^3 + (11664*p^6 + 157464*p^3*q^2 + 531441*q^4)*d);
> w:=(1/9*d^6*e + 27/2*q*d^6 + 2*p*d^5*e + (-12*p^3 - 81*q^2)*d^3*e +
> (1458*p^3*q + 19683/2*q^3)*d^3 + (-432*p^4 - 2916*p*q^2)*d^2*e +
> (-2592*p^6 - 34992*p^3*q^2 - 118098*q^4)*e)/(d^7
> + 9*p*d^6 + 81*p^2*d^5 + (216*p^3 + 1458*q^2)*d^4 + (972*p^4 +
> 6561*p*q^2)*d^3 + (11664*p^6 + 157464*p^3*q^2 + 531441*q^4)*d);
> m:= e*( d^5 + 12*p*d^4 + 54*p^2*d^3 + (108*p^3 + 729*q^2)*d^2 +
> (-648*p^4 - 4374*p*q^2)*d) +
> 81/2*q*d^5 + (4374*p^3*q + 59049/2*q^3)*d^2;
> n:=d^6 + 9*p*d^5 + 81*p^2*d^4 +
> (216*p^3 + 1458*q^2)*d^3 + (972*p^4 + 6561*p*q^2)*d^2 +
> (11664*p^6 + 157464*p^3*q^2 + 531441*q^4);
> phi:=map<E-> C | [u, 3*u+3*m/n-3*w]>;
> time Image(phi) eq C;
true
Time: 66.680
If we pick the point $P:=(-3p,\frac{27}2 q)$ on $E$, it can be verified that the preimage of the point $\varphi(nP) \in C$ has cardinality $1$ for all $n\in \mathbb Z$ such that $|n|<5$. It thus seems plausible that the map $\varphi : E \to C $ is in fact a birational isomorphism.
Question 1. Is it possible to find the inverse of the map $\varphi$?
Question 2. How can we transform the curve $C$ into Weierstrass normal form without knowing about $E$ in advance?
