Solve $a^3+b^3+3ab=1$ with $(a,b)\in \Bbb{Z}^2$ 
Solve the following equation for $(a,b)\in \Bbb{Z}^2$:
  $$a^3+b^3+3ab=1$$

I tried all of the standard techniques I know. I tried modular arithmetic:
$$a^3+b^3+3ab\equiv 1 \pmod{3} $$
$$a^3+b^3\equiv 1 \pmod{3} $$
Now by Fermat's Little Theorem:
$$a^2 a+b^2 b\equiv 1 \pmod{3} $$
$$a+b\equiv 1 \pmod{3} $$
But I can't see the next move I have to do. I can't find any banal factorization of the first term(it would require solving a $3$ degree equation). I tried using classic scomposition such as the sum of $2$ cubes and the cube of a binomial. Thank you for your time :)

The item worth memorizing is
$$ x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( x^2 + y^2 + z^2 - yz - zx - xy \right) $$
where the quadratic form is positive semidefinite because
$$ \left( x^2 + y^2 + z^2 - yz - zx - xy \right) = \frac{1}{2} \left( (y-z)^2  + (z-x)^2 + (x-y)^2 \right) $$
If you then take $z=-1$ you get
$$ x^3 + y^3 -1 + 3xy = (x+y-1)\left( x^2 + y^2 +1 + y + x - xy \right) $$
and the quadratic factor is
$$  \frac{1}{2} \left( (y+1)^2  + (-x-1)^2 + (x-y)^2 \right)=  \frac{1}{2} \left( (y+1)^2  + (x+1)^2 + (x-y)^2 \right) $$
