Birational equivalence of cubic with a Weierstrass form I want to convert the cubic $x^3 + 3y^3 - 11z^3 = 0$ to Weierstrass form (to find its rank) so I tried to follow the suggestion from Timothy: I found three points $(2:1:1),(28:-19:5),(-537656:443213:212645)$ on the curve and used a linear change of variables to map them to $(1:0:0),(0:1:0),(0:0:1)$ then I dehomogenized and set $y' = xy$ then homogenized that to get $1330 x'^3 - 252013629 x'^2 z' = y'^2 z' - 2352637000 y' z'^2 - 56454 x' y' z'$. It doesn't seem to be birationally equivalent to what I started with (due to the rehomogenization), $(1:0:0)$ isn't a zero anymore (but $(1:0:9480240/201019314751)$ is).
So if anyone could tell me where I went wrong, or suggest a different technique or reference on how to convert elliptic curves into Weierstrass form that would be a big help. Thanks.
 A: This answer is most likely too late, but here it is just in case others find it useful. The general procedure to transform a genus $1$ cubic projective curve into Weierstrass form is outlined in Silverman/Tate "Rational points on elliptic curves", Chapter I, Section 3, pages 22-23.
Some software packages can do this for you. For instance, Magma can do it. Magma has an online calculator (which is free to use as long as the calculation does not go over 60 seconds). Here is some code that will take your curve and output a Weierstrass form, a minimal model for the curve, and the rank:
PP<x,y,z>:=ProjectiveSpace(Rationals(),2);
C:=Curve(PP,x^3+3*y^3-11*z^3);
P0:=C![2,1,1];
E, phi:=EllipticCurve(C,P0);
Em, psi:= MinimalModel(E);
E;
Em;
Rank(Em);

The output that we obtain is:
Elliptic Curve defined by y^2 = x^3 + 7776/25*x^2 + 20155392/625*x -
181398528/625 over Rational Field
Elliptic Curve defined by y^2 + y = x^3 - 7351 over Rational Field
1

So your curve is given by $y^2+y=x^3-7351$ and its Mordell-Weil rank is $1$.
