74th Romanian Mathematical Olympiad

The equality can be written $a^4{b}^4-3a^6-3b^6+9a^2{b}^2=0$, $a^4(b^4-3a^2)-3b^2(b^4-3a^2)=0$, $(a^4-3b^2)(b^4-3a^2)=0$. So $a^4=3b^2$ or $b^4=3a^2$ and, since $a\ne 0$ and $b\ne 0$, $\sqrt{3}=\pm \frac{a^2}{b}$ or $\sqrt{3}=\pm \frac{b^2}{a}$. If $a$ and $b$ are rational, then so is $\sqrt{3}$, false. So the assumption is false, hence at least one of the numbers $a$ and $b$ is irrational.