A cubic diophantine Equation While reading Diophantine equations I came across the following equation $$x^3+cy^3-3yx=0$$
Is there any known method to solve this equation for any $c$?
 A: We have $y\mid x^3$ and $x\mid cy^3$. If we assume $y=ax$ for some integer $a$ then
substituting this in $x^3+cy^3=3xy$ yields
$$
x=a(3-ac), \quad y=a^2(3-ac).
$$
Conversely this is a solution of $x^3+cy^3=3xy$ for all integers $a$. This is a first step to solve this Diophantine equation (which is not homogeneous, i.e., not given by a binary cubic).
A: First note that the solutions involving zero are $x=y=0$ or $c=0,x=3t,y=3t^2$. Otherwise, we can give full solutions in terms of the (rare) solutions of much simpler equations. 

For $c\ne0$, the equation $x^3+cy^3=3xy$ is solvable in non-negative integers if and only if there is a solution in non-negative integers of the equation $$a^2u^3+bv^3=1\text{ or }3,\text{ where }c=ab.$$

Proof
Let $x=tz,y=tv$, where $z$ and $v$ are corime. Then $$t(z^3+cv^3)=3zv.$$
$z$ is a factor of $tcv^3$ and therefore of $tc$. Let $c=ab$ and $z=au$, where  $b$ and $u$ are coprime and where we can choose $b$ to be positive. Then $$t(a^2u^3+bv^3)=3uv.$$
Then $uv$ is coprime to $a^2u^3+bv^3$ and so $t=suv$ and  $$s(a^2u^3+bv^3)=3.$$
If this equation is solvable then the original equation has solution $x=asu^2v,y=suv^2.$ Note that switching the signs of $s,u$ and $v$ leaves $x$ and $y$ unchanged so we can suppose $s$ is positive.
Example 1  Solve $x^3+y^3=3xy.$
The equation $u^3+v^3=1$ has no non-zero solutions (Fermat) and the equation $u^3+v^3=3$ is impossible modulo $9$. So $x^3+y^3=3xy$ has no non-zero solutions. 
Example 2  Solve $x^3+6y^3=3xy.$
The equation $4u^3+3v^3=1 (s=3)$, for example, has solution $u=1,v=-1$, giving $x=-6,y=3.$
In general one can use PARI/GP to obtain all solutions. For example see the solution by @YongHaoNg to the following post:- 
Solutions of $ax^3+by^3=1$
A: first we have (0,0,0),(0,a,0),(0,0,c) are solutions ∀ y=a  ,a,c∈Z
now,
by multiplying by xy we have
$yx^4-3x^2 y^2+cxy^4=0$
let,
$x^2=s   ,y^2=r ,   y=a   ,cx=k$
then 
$as^2-3sr+kr^2=0$
the discriminant  ,     $d=9-4ak=9-4cxy$
as d should be greater than or equal to zero and for Diophantine equations d is a perfect square 
$d>=0  , d=u^2$ 
$cxy<=2$, if we considered that cxy>=0
implies  $cxy=2$ , or $cxy=0$  because 1 makes d non-square
so both x,y >=0  or x,y<=0 together
take c=∓2 ,
$xy=∓1,  x^3∓2y^3∓3=0 ,(∓1,1) -is-the-only-solution $
take c=∓1 ,
$xy=2,  x^3∓y^3=∓6 , -has-no-solution $
take c=0 ,
$  x^3-3xy=0 , -has-infinite-solutions $
if cxy<0 ,
$  x^3+cy^3-3xy=0 , -has-infinite-solutions $
hint: d=u^2 and d is odd then if cxy<0    2-cxy=m(m-1), $m∈z^+$
(m-2)(m+1)=-cxy  ,∀ m>2 & either one of c,x,y is negative or all
