= Diagonal cubic elliptic curve
{title2=$X^3+Y^3+dZ^3=0$}
For $d\in\mathbb Q^\times$, the <smooth projective curve> $X^3+Y^3+dZ^3=0$, with identity $O=(1:-1:0)$, is a genus-one <elliptic curve>. Its partial derivatives have no common projective zero. The invertible linear <projective transformation>
$$
(\mathcal X:\mathcal Y:\mathcal Z)=(-12dZ:-36d(X-Y):X+Y)
$$
sends it to $\mathcal Y^2\mathcal Z=\mathcal X^3-432d^2\mathcal Z^3$, taking $O$ to $(0:1:0)$. Indeed $(X+Y)^3+3(X+Y)(X-Y)^2=4(X^3+Y^3)=-4dZ^3$. 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>.
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