All pairs (x,y) that satisfy the equation $xy+(x^3+y^3)/3=2007$

How we can find the all pairs $(x,y)$ from the integers numbers ,that satisfy the equation :

$$xy+\frac{x^3+y^3}{3} =2007$$

Observe that the equation is symmetric.

As $3|(x^3+y^3)$, either $(x,y)$ will be $(3a+1,3b-1)$, $(3a-1,3b+1)$ or $(3a,3b)$.

If $(x,y)$ is $(3a+1,3b-1)$, $\frac{x^3+y^3}{3}=3(3a^3+3b^3+3a^2-3b^2+a+b)$

So, 3 must divide $xy$ which is impossible as $xy=(3a+1)(3b-1)$

So, $(x,y)$ will be $(3a,3b)$.

So,$9(ab+a^3+b^3)=2007\implies a^3+b^3+ab=223$

Now, 223 is prime, so, $(a,b)=1$

If we think of solution in natural number, $a<7$ .

By trial (which is aided by $(a,b)=1$), $(a,b)$ is $(6,1)$ or $(1,6)$.

Let $x+y=a,xy=b$ then the equation is equivalent to $$a^3-3abc+3b=6021$$ or $$(a-1)(a^2+a+1-3b)=6020$$ Now it is easy to do.