Let $$\theta=\sqrt[3]2,K=Q(\theta),R=Z[\theta],$$ where R is the ring of integers (has class number 1).I was trying to solve the diophantine equation $x^3+2y^3=3z^3,$ which becomes equivalent to $$x+y\theta=\epsilon.\lambda.\alpha^3,$$ where $\epsilon$ is a unit and $\lambda$ is the prime $1+\theta$ (we need only consider when the unit is 1, since $1+\theta+\theta^2=\frac{\lambda^3}{3}$) .In a paper by Franz Lemmermeyer,particularly the sixth section, it is proved that the classes of so called "binomial" cubes (their $\theta^2$ component is 0) are in bijection with rational points of an elliptic curve.Can we do something similar to the equation above, if we substitute $$\alpha=A+B\theta+C\theta^2?$$Would this be enough to solve the equation in relatively prime integers x,y,z?It becomes equivalent to the following:
$$A^2B+A^2C+B^2A+2B^2C+2C^2A+2C^2B=0,$$ which is homogenous, so we can divide by the cube of one of the three.I am stuck here, though.
Note: a particular paper by T. Nagell seems relevant, but I cannot access it.It is called "Solution complète de quelques équations cubiques à deux indeterminées." Journal de Mathématiques Pures et Appliquées 4 (1925): 209-270.
HINT.-There are infinitely many rational solutions because of the equivalence (very known on elliptic curves $X^3+Y^3=AZ^3$) $$x+y=z\iff(9xy^2+(y-x)^3)^3+(9x^2y-(y-x)^3)^3=xyz(3(x^2+xy+y^2))^3\qquad (*)$$
In fact, $(1,1,1)$ is an obvious solution of $x^3+2y^3=3z^3$ so we have $$1+2=3\iff (9\cdot1\cdot4+1)^3+(9\cdot1^2\cdot2-1)^3=6(3(1+2+2^2))^3$$ i.e. $$1+2=3\iff37^3+17^3=6(21)^3\qquad(**)$$ Our obvious solution $(1,1,1)$ corresponds to the generator $G_0=(37,17,21)$ of the elliptic curve $X^3+Y^3=6Z^3$ which is known has rank equal to $1$ hence all its rational points are given by $nG_0$ where $nG_0=\underbrace{G_0+G_0+\cdots +G_0}_{n\text{ times}}$ and $+$ is the sum of the group all elliptic curve is.
In order to find possible integer solutions of $x^3+2y^3=3z^3$ other than $(1,1,1)$ we can calculate $3nG_0$ and compare the resulting expression using $(*)$. However this requires calculations of degree polynomials $ (3n)^2$, in particular for $3G$ we should use degree $9$. Calculations of rational points with elliptic curves always involves big integers (Who wanted to do it can use the formulas for the addition on $X^3+Y^3=AZ^3$. There are infinitely many rational solutions but only finitely many integer solutions and maybe just one, the $(1,1,1)$.
Historic note (**).-Legendre in his "Essai sur la théorie des nombres" (Second edition 1808) erroneously added $6$ to his first $2,3,4$ and $5$ not representable as sum of two rational cubes.