 Diagonal cubic elliptic curve

For $d\in {\mathbb{Q}}^{\times }$, the smooth projective curve $X^3+Y^3+d{Z}^3=0$, with identity $O=(1:-1:0)$, is a genus-one elliptic curve. Its partial derivatives have no common projective zero. The invertible linear projective transformationsends it to ${\mathcal{Y}}^2\mathcal{Z}={\mathcal{X}}^3-432d^2{\mathcal{Z}}^3$, taking $O$ to $(0:1:0)$. Indeed $(X+Y{)}^3+3(X+Y)(X-Y{)}^2=4(X^3+Y^3)=-4d{Z}^3$. In affine coordinates $u=X/Z,v=Y/Z$, this gives $x=-12d/(u+v)$ and $y=-36d(u-v)/(u+v)$, with inverse $u=(y-36d)/(6x)$, $v=(-y-36d)/(6x)$. Thus diagonal sums of two cubes are naturally studied through elliptic-curve addition formula.