I would like to show that the following result holds:
The equations $x^3+y^3=1$ and $y^2=4x^3-1$ are the same Riemann surface in $\mathbb{CP}^2$ and as a consequence there are two meromorphic functions $f,g$ such that $f^3+g^3=1$.
I first tried using Able's theorem but didn't get very far using this approach.
Using compactification I managed to find the points that needed to be addaad to the surface of each of the equations but I don't how to proceed with this attempt in order to complete my proof of the desired claim.
I'm thinking if this proof is correct or not:
$x^3+y^3=1$ is the Riemann surface in $\Bbb{CP}^2$ define by:
$$ \{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\} $$
Let
$$ \begin{aligned} X &= U \\ Y &= \frac{\sqrt{3}}{6}V -\frac{1}{2}W \\ Z &= \frac{\sqrt{3}}{6}V +\frac{1}{2}W \\ \end{aligned} $$ Then $(X, Y, Z) \mapsto (U, V, W)$ is a biholomorphic automorphism in $\Bbb{CP}^2$ and:
$$ \begin{aligned} X^3+Y^3-Z^3&=U^3+(\frac{\sqrt{3}}{6}V-\frac{1}{2}W)^3-(\frac{\sqrt{3}}{6}V+\frac{1}{2}W)^3 \\ &=U^3-\frac{1}{4}W(V^2+W^2) \end{aligned} $$ Hence $$ \{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\} \\ = \{[U, V, W] \in \Bbb{C}^3|U^3-\frac{1}{4}W(V^2+W^2)=0, (U, V, W) \neq (0, 0, 0)\} $$
Setting $\frac{U}{W}=u, \frac{V}{W}=v$, in affine coordinate, it's $v^2=4u^3-1$.