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For $Y^{2}Z=X(X-Z)(X-\lambda Z)$ over $\mathbb{C}$, order its three finite roots as $(r_0,r_1,r_2)$ and put $h=r_1-r_0$. The projective transformation $X^{\prime }=(X-r_{0}Z)/h$, $Y^{\prime }=Y/s$, $Z^{\prime }=Z$, with $s^2=h^3$, gives Legendre form of an elliptic curve with parameter $(r_2-r_0)/(r_1-r_0)$. The six root orderings give the six displayed values, all different from zero and one. These exhaust changes that act affinely on $X/Z$ and scale $Y/Z$, because such a change must send the finite root set onto the new finite root set. There are six distinct values except that $\lambda \in \{-1,2,1/2\}$ gives three and ${\lambda }^2-\lambda +1=0$ gives two.

First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex $[0:0:1]$ to the point at infinity $[0:1:0]$ of a Weierstrass model.

Put $h=\omega -1$ and choose $s\in \mathbb{C}$ with $s^2=h^3$. Both are nonzero. DefineSince $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, $\lambda \ne 0,1$.

Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is $-\sqrt{3}/(hs)\ne 0$, so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.

Choose a flex as $O=[0:1:0]$, with tangent line $Z=0$. The flex coordinates for a smooth plane cubic giveThe coefficient $a$ is nonzero because $Z=0$ is not a component, and $b\ne 0$ expresses smoothness at $O$. Since the field characteristic is not two, the projective linear changecompletes the square. The equation becomes $Y^{\prime 2}Z=P_3(X,Z)$ after dividing by a nonzero constant, where $P_3$ is a degree-three homogeneous polynomial with nonzero $X^3$ coefficient.

On $Z=1$, the polynomial $P_3(x,1)$ has three distinct roots $r_1,r_2,r_3$ in the algebraically closed field. A repeated root $r$ would make $(r,0)$ a singular point of $y^{\prime 2}=P_3(x,1)$, since both first partial derivatives vanish there. PutA further nonzero scaling of $y^{\prime }$ absorbs the leading coefficient and $(r_2-r_1{)}^3$; the required square root exists in the algebraically closed field. These are projective linear changes of the original coordinates. Renaming the new coordinates gives the Legendre form of an elliptic curveIts projective completion is $Y^{2}Z=X(X-Z)(X-\lambda Z)$, with the chosen flex at infinity.

For the Legendre form of an elliptic curve $y^2=x(x-1)(x-\lambda )$, a change in the $x$ coordinate that permutes its branch values lifts with either of two opposite signs in the $y$ coordinate. These lifts differ by the involution $(x,y)\mapsto (x,-y)$. At $\lambda =e^{\pi i/3}$ the parameter transformation $x\mapsto 1-x/\lambda$ has order three. Its lift fixing $y$ has order three, whereas its lift negating $y$ has order six, since its cube is the nontrivial involution. Thus the order must be computed for a specified lift; both imply the same hexagonal shape of the associated period lattice.