How to solve $x^3 + 2y^3 + 3z^3 = 4xyz$ for $x,y,z \in \mathbb{Q}$ solve $x^3 + 2y^3 + 3z^3 = 4xyz$ for $x,y,z \in \mathbb{Q}$
My main question is how to solve this in $\mathbb{Q}$, i know how to solve these kind of problems in $\mathbb{Z}$, where i usually look at modulo small primes, such as 3,5. But with $\mathbb{Q}$ i get totally confused, any smart methods i can use solving this?
Kees
 A: If integer $X,Y,Z$ are solutions of $$X^3+2Y^3+3Z^3=4XYZ,\tag{1}$$
then for any integer $D\ne 0$ one will have
$$
\left(\frac{X}{D}\right)^3 +
2\left(\frac{Y}{D}\right)^3 +
3\left(\frac{Z}{D}\right)^3 =
4\frac{X}{D}\frac{Y}{D}\frac{Z}{D}.
$$
So, one will have rational solution $(x,y,z)=\left(\frac{X}{D},\frac{Y}{D},\frac{Z}{D}\right)$.
But equation $(1)$ has no nontrivial solutions.
If it will have nontrivial solution $(X,Y,Z)$, then denote $G = GCD(X,Y,Z)$.
Then denote $X_0=X/G$, $Y_0=Y/G$, $Z_0=Z/G$, and $(X_0,Y_0,Z_0)$ will be solution of $(1)$ too.
Now, consider all by modulo 7.
Since $GCD(X_0,Y_0,Z_0)=1$, then they all cannot be divisible by $7$.
$0^3\equiv 0 (\bmod 7)$,
$1^3\equiv 1 (\bmod 7)$,
$2^3\equiv 1 (\bmod 7)$,
$3^3\equiv -1 (\bmod 7)$,
$4^3\equiv 1 (\bmod 7)$,
$5^3\equiv -1 (\bmod 7)$,
$6^3\equiv -1 (\bmod 7)$.
Considering all possible values of  $X_0,Y_0,Z_0 (\bmod 7)$ in the $\{0,1,2,3,4,5,6\}^3$, one will get only solution $(0,0,0)$, which mean that $X_0,Y_0,Z_0$ all are divisible by $7$. So, only one integer solution exists: $(0,0,0)$.
