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 form an *affine plane algebraic curve*, which can be constructed as follows. First, consider the variety $ V$ lieing 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$ given by $ (u,v,x_1,x_2,x_3,y_1,y_2,y_3) \mapsto (u,v)$ . We can get an explicit equation for $ 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 that although the map $ f$ is not invertible, the bijection between rational solutions of equation $ (1)$ and rational points on the curve $ C$ is obvious: $ (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 a birational isomorphism; hence the **questions**:

- Is it possible to find the inverse of the map $ \varphi$ ?
- How can we transform the curve $ C$ into Weierstrass normal form without knowing about $ E$ in advance?