58th Ukrainian National Mathematical Olympiad

Rewrite given equation as $a^6-7a^3{b}^3-8b^6=0$, hence $(a^3+b^3)(a^3-8b^3)=0$. We can show that $x^2\pm xy+y^2>0$ holds iff $x,y\ne 0$, since $$x^2\pm xy+y^2=(x\pm \frac{1}{2}y{)}^2+\frac{3}{4}{y}^2=0\iff y=0\text{ and }x\pm \frac{1}{2}y=0\iff x=y=0.$$

Hence given equation can be written as $a^3+b^3=0$ or $a^3-8b^3=0$. And at the same time values $a,b$ cannot both be zero.

Case 1. $a^3+b^3=0\Rightarrow (a+b)(a^2-ab+b^2)=0\Rightarrow a=-b$, since $a^2-ab+b^2\ne 0$. Hence $$\frac{a^2-b^2}{a^2+b^2}=\frac{b^2-b^2}{b^2+b^2}=0.$$

Case 2. $a^3-8b^3=0\Rightarrow (a-2b)(a^2+2ab+4b^2)=0\Rightarrow a=2b$, since $a^2+2ab+4b^2\ne 0$. Hence $$\frac{a^2-b^2}{a^2+b^2}=\frac{4b^2-b^2}{4b^2+b^2}=\frac{3}{5}.$$