How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that 
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?

The first thing that screams at me is $3|x^3 + x^3$ so
$x^3 \equiv -y^3 \mod 3$ so $x\equiv -y \mod 3$
Let $x \equiv i \mod 3$ and $y \equiv -i \mod 3$.  If $x = 3k + i$ and $y = 3j -i$ then $\frac {x^3 + y^3}{3} = 9(k^3 + j^k) + 9(k^2i-j^2i)+3(ki -ji) + \frac {i^3 - i^3}3 \equiv 0 \mod 3$
So $xy + \frac  {x^3 + y^3}3 \equiv i^2 \mod 3$ and $2007\equiv 0 \mod 3$ so $i = 0$ and $3|x$ and $3|y$.
Let $x = 3a$ and $y = 3b$ and we get
$9ab + 9(a^3 + b^3) = 2007$ so 
$ab + a^3 + b^3 = 223$
Hmmm, still trial and error but $b^3, a^3 < \sqrt[3]223 \approx 6$ so not too many to test.  
One of $a$ or $b$ must be odd. We have $ab + a^3 + b^3\equiv ab + a +b \equiv 1 \mod 3$:  That means:
$(a,b)\equiv (0, k) \to 0*k + 0 + k\equiv k \mod 3$
So $a\equiv 0$ (wolog) and $b \equiv 1$ is possible.
$(a,b) \equiv (1,1) \to 1 + 1 + 1 \equiv 0 \mod 3$. 
$(a,b) \equiv (1,-1) \to -1 + 1 + -1 \equiv -1 \mod 3$.
$(a,b) \equiv (-1,-1) \to 1 -1 -1 \equiv -1 \mod 3$.
So $a \equiv 0$ and $b \equiv 1$ and at least one is odd.
So 
$[0,1],[3,1],[3,4],[6,1]$ are the only four options. 
Obviously if the average of $ab, a^3, b^3$ is $\frac {223}3 \approx 70 > 4^3$, the first three can't possibly work.  Don't even need to test them.
So $[6,1]$ is only option.
An indeed $6 + 6^3 + 1 = 223$.
So, if my reasoning is right (which.... I really don't see any error or faulty assumptions... but ... a self-editor is always blind....).  $\{x,y\} = \{18,3\}$ are the only solutions.
