Slovenian Mathematical Olympiad, 2016

Adding $3y^3$ to the equation and factoring both sides we obtain $4(x + y)(x^2 - xy + y^2) = 3y^2(x + y)$. Moving all terms to the left side we can bring out $(x + y)$, so $(x + y)(4x^2 - 4xy + y^2) = 0$. By factoring the second expression we get $(x + y)(2x - y)^2 = 0$. So, $y = -x$ or $y = 2x$. Since $x$ and $y$ are relatively prime, we conclude that $x$ can only be $1$ or $-1$. From this we get four pairs $(x, y)$ of relatively prime solutions to the equation, namely $(1, -1)$, $(-1, 1)$, $(1, 2)$ and $(-1, -2)$.