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 $(1,6)$.
