Homogeneous spaces of elliptic curve Let $d>1$ be a cube free natural number  and $a,b,c$ are natural numbers greater than 1 with $abc=d$. How to explain that the curve $D: ax^3+by^3+cz^3=0$ is homogeneous space (over $\mathbb{Q}$) for the curve $E:x^3+y^3+dz^3=0$?
First: to prove that $D\in Twist(E/\mathbb{Q})$ is sufficient to construct $\overline{\mathbb{Q}}$-isomorphism $\phi:E\to D$, but how to do this?
Second:  to show that $D\in WC(E/\mathbb{Q})$ is sufficient to construct (with $\phi$) 1-cocycle corresponding to $D$ in $H^1(G,Isom(E))$ such that it takes values in the group of translations of E. (G is absolute Galois group of $\mathbb{Q}$, $WC(E/\mathbb{Q}$) is Weil-Chatelet group; $Twist(E/\mathbb{Q})$ is the set of twistor spaces of E). 
 A: The following is an idea based on Cassel's Lectures on Elliptic Curves (see chapter 20, exercise 3).
Consider the real third-root of $a$, say $a^{1/3}$ (the other roots are $\rho{a^{1/3}}$ and $\rho^{2}{a^{1/3}}$ where $\rho$ is a primitive third-root of unity). Define $\phi:C\rightarrow{E}$ by sending $(x,y,z)$ to $(a^{1/3}x,b^{1/3}y,\frac{z}{a^{1/3}b^{1/3}})$. You can check that the inverse $\phi^{-1}:E\rightarrow{C}$ is given by sending $(x,y,z)$ to $(\frac{x}{a^{1/3}},\frac{y}{b^{1/3}},a^{1/3}b^{1/3}z)$, so that $\phi$ is a $\overline{\mathbb{Q}}$-isomorphism. 
If $\sigma\in{\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})}$, then note that $\phi^{\sigma}(x,y,z)=((a^{1/3})^{\sigma}x,(b^{1/3})^{\sigma}y,\frac{z}{(a^{1/3}b^{1/3})^{\sigma}})$ and note that $(a^{1/3})^{\sigma}$ can take the values $a^{1/3}$, $\rho{a^{1/3}}$ or $\rho^{2}{a^{1/3}}$. Something similar happens in the cases $(b^{1/3})^{\sigma}$ and $(a^{1/3}b^{1/3})^{\sigma}$.
Then you get $\phi^{\sigma}\phi^{-1}(x,y,z)=(\frac{(a^{1/3})^{\sigma}}{a^{1/3}}x,\frac{(b^{1/3})^{\sigma}}{b^{1/3}}y,\frac{a^{1/3}b^{1/3}}{(a^{1/3}b^{1/3})^{\sigma}}z)$. Suppose, for example, that $(a^{1/3})^{\sigma}=a^{1/3}$ and $(b^{1/3})^{\sigma}=b^{1/3}$. Then $\phi^{\sigma}\phi^{-1}(x,y,z)=(x,y,z)$, so you get the identity on $E$, which you can view as the translation map by the origin $O\in{E}$ (the curve $E$ can be viewed as an elliptic curve over $\mathbb{Q}$). By working out the other possible values of $(a^{1/3})^{\sigma}$ and $(b^{1/3})^{\sigma}$, you can verify that $\phi^{\sigma}\phi^{-1}$ is either a translation by the point $(\rho,-\rho^{2},0)$ or by the point $(\rho^{2},-\rho,0)$, in any case you obtain that $\phi^{\sigma}\phi^{-1}$ is a translation map.  
