Rational Parametrization of cubic curve $x^3+y^3=1$ How to rationally parametrize the cubic curve $x^3+y^3=1$? It's a slice of Fermat cubic, and the Fermat cubic does have rational parametrization, so I think it should also have rational parametrization.
I tried $u=x+y, v=x-y$ and got $u(u^2+3v^2)=4$,or $v=\pm\sqrt{\frac{4-u^3}{3u}}$, which is a bit closer to what I want, but still not good enough.
 A: HINT.- You can do it however it is not "elementary".
The elliptic curve $E_1:X^3+Y^3=1$ can be rationally parametrized but with a trascendental parameter, a (corresponding) Weierstrass function $\wp (z)$. In order to do this, you can first put $V_1$ under its Weierstrass form which is $$E_2:y^2=x^3-432$$ which comes from the birational transformation $$(X,Y)\rightarrow \big(\frac{12}{X+Y},\frac{36(X-Y)}{X+Y}\big)$$ whose reciprocal is $$(x,y)\rightarrow\big(\frac{36+y}{6x},\frac{36-y}{6x}\big)$$
Now you have the classical formula linking $\wp(z)$ and its derivative $\wp'(z)$ 
$$\wp'^2=4\wp^3-g_2\wp-g_3$$ where $g_2$ and $g_3$ are the invariants of $\wp$ which must be calculated $$g_2=60\sum_{\omega\ne 0}\frac{1}{\omega^4}\text{ and}\space g_3=140\sum_{\omega\ne 0}\frac{1}{\omega^6}$$ where $\omega\in L$. 
($L$ is the corresponding lattice for the  doubly periodic function $\wp(z)$ which if I remember correctly is hexagonal).
This way we have the parametrization $$\begin{cases}x=\wp(t)\\y=\wp'(t)\end{cases}$$ 
A: For $x\geq0$ and $y\geq0$ we have $x=\sqrt[3]{\sin^2\theta}$ and $y=\sqrt[3]{\cos^2\theta}$.
The similar for $x\leq0$ and $y\leq0$.
In the rest cases we can use $\sinh$ and $\cosh$.
