Solve in integers the equation  
$$ x^{2}+x y+y^{2}=\left(\frac{x+y}{3}+1\right)^{3} $$
We do the trick of setting $a=x+y$ and $b=x-y$. This rewrites the equation as 
$$ \frac{1}{4}\left((a+b)^{2}+(a+b)(a-b)+(a-b)^{2}\right)=\left(\frac{a}{3}+1\right)^{3} $$
where $a, b \in \mathbb{Z}$ have the same parity. This becomes 
$$ 3 a^{2}+b^{2}=4\left(\frac{a}{3}+1\right)^{3} $$
which is enough to imply $3 \mid a$, so let $a=3 c$. Miraculously, this becomes 
$$ b^{2}=(c-2)^{2}(4 c+1) $$
$$ x=\frac{1}{8}\left(3\left(m^{2}-1\right) \pm\left(m^{3}-9 m\right)\right) \quad \text { and } \quad y=\frac{1}{8}\left(3\left(m^{2}-1\right) \mp\left(m^{3}-9 m\right)\right). $$
For mod 8 reasons, this always generates a valid integer solution, so this is the complete curve of solutions. Actually, putting $m=2 n+1$ gives the much nicer curve 
$$ x=n^{3}+3 n^{2}-1 \quad \text { and } \quad y=-n^{3}+3 n+1 $$
and permutations.