Problem: Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]

Solution: We do the trick of setting $a=x+y$ and $b=x-y$.
This rewrites the equation as
\[ \frac14\left( (a+b)^2+(a+b)(a-b)+(a-b)^2 \right) = \left( \frac a3 + 1 \right)^3 \]
where $a,b \in \ZZ$ have the same parity.
This becomes
\[ 3a^2+b^2 = 4\left( \frac a3 + 1 \right)^3 \]
which is enough to imply $3 \mid a$, so let $a = 3c$.
Miraculously, this becomes \[ b^2 = (c-2)^2 (4c+1). \]

So a solution must have $4c+1=m^2$, with $m$ odd.
This gives
\[ x = \frac 18 \left( 3 (m^2-1) \pm (m^3-9m)  \right)
  \quad\text{and}\quad
  y = \frac 18 \left( 3 (m^2-1) \mp (m^3-9m) \right).  \]
For mod $8$ reasons,
this always generates a valid integer solution,
so this is the complete curve of solutions.
Actually, putting $m=2n+1$ gives the much nicer curve
\[ \boxed{x = n^3+3n^2-1 \quad\text{and}\quad y = -n^3+3n+1} \]
and permutations.

For $n=0,1,2,3$ this gives the first few solutions
are $(-1,1)$, $(3,3)$, $(19,-1)$, $(53, -17)$, (and permutations).