<s>[INST] Find pairs of $(x, y)$ satisfying the given equation. Find all pairs $(x, y)$ of integers such that $$xy+\frac{x^3+y^3}{3}=2007$$
$\mathbf{Try:}$
The given expression can be simplified as $$x^3+y^3+3xy=6021$$
$$\Rightarrow (x+y-1)[(x-y)^2+(y+1)^2+(x+1)^2]=12040$$
I tried considering the divisors of $12040$ but seems to be a much tedious task. Now how to proceed further. 
 [/INST] $$\frac{x^3+y^3}{3}=\frac{(x+y)^3-3xy(x+y)}{3},$$ which says that $(x+y)^3$ is divided by $3$ and from here $x+y$ is divided by $3$, 
but it gives that $xy$ is divided by $3$.
Thus, $x$ and $y$ are divided by $3$.
Let $x=3a$ and $y=3a$, where $a$ and $b$ are integers.
Hence, we obtain $$ab+a^3+b^3=223.$$
It's a bit of easier. 
Also, we have
$$(x+y-1)(x^2+y^2+1-xy+x+y)=6020=4\cdot5\cdot7\cdot43$$
and $x+y-1\equiv2(\mod3),$ which gives not so many cases because $$x^2+y^2+1-xy+x+y>0.$$
Now, it's obvious that $x+y>0$ and
$$6020=(x+y-1)(x^2+y^2-xy+x+y+1)=(x+y-1)((x+y)^2-3xy+1)\geq$$
$$\geq(x+y-1)\left((x+y)^2-\frac{3}{4}(x+y)^2+x+y+1\right)=\frac{1}{4}(x+y-1)(x+y+2)^2,$$
which gives
$$x+y\leq27.$$
We see that $x+y-1=14$ or $x+y-1=20$ only.
If $y=15-x$ then $$x^2+(15-x)^2+1+15-x(15-x)=430,$$
which has no integer roots.
If $y=21-x$ then
$$x^2+(21-x)^2+1+21-x(21-x)=301$$ or
$$(x-3)(x-18)=0,$$
which gives the answer:
$$\{(3,18),(18,3)\}.$$
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