Isomorphism via birational equivalence
I'm looking at Exercise 1.13 in Rational Points of Elliptic Curves by Silverman and Tate.
I know, from Part (a) of the question, that if $u$ and $v$ satisfy the relation $u^3 + v^3 = \alpha$, then the quantities $$x = \frac{12\alpha}{u+v} \;\;\;\; \text{and} \;\;\;\; y=36\alpha\frac{u-v}{u+v}$$ satisfy the relation $y^2 = x^3 - 432\alpha^2$, giving a birational transformation from the curve $u^3 + v^3 = \alpha$ to the curve $y^2 = x^3 - 432\alpha^2$. How would we prove that the birational transformation described in Part (a) is an isomorphism of groups?
I'm wondering how to approach this. Do I need to set up the map, say $\phi$, between the two curves and then show that $\phi(P+Q) = \phi(P) + \phi(Q)$?
Any guidance would be much appreciated!
You can actually prove it in a more conceptual way, and here are some key steps:
Denote by $E_1$ and $E_2$ the curves $ u^3+v^3=\alpha w^3$ and $y^2z=x^3 -432\alpha^3z^3$. The morphism in (a) is then given by $\phi\colon[u:v:w]\mapsto [12\alpha w,36\alpha (u-v),u+v]$. After choosing $[1,-1,0]\in E_1$ and $[0,1,0]\in E_2$ as identities and endowing the natural group structure on them, both $E_1$ and $E_2$ become elliptic curves.
Every birational map between elliptic curves is an isomorphism, so is $\phi$;
Since $\phi$ sends identity to identity, such a map must be a group homomorphism.