## Task A-1.2.

Let $a$ and $b$ be positive real numbers such that $a^{3}+b^{3}=2 a b(a+b)$. Determine $\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}$.
## Solution.

By factoring the left side, we get $(a+b)\left(a^{2}-a b+b^{2}\right)=2 a b(a+b)$.

Since the numbers $a$ and $b$ are positive, it follows that $a+b \neq 0$, so we can divide the obtained equation by $a+b$.

Therefore, it follows that $a^{2}-a b+b^{2}=2 a b$, or $a^{2}+b^{2}=3 a b$.

By dividing by $a^{2} b^{2} \neq 0$, we get $\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=\frac{a^{4}+b^{4}}{a^{2} b^{2}}=7$.

Note: It is necessary to write that $a+b$ and $a b$ are not zero. If the student does not mention the possibility of division by zero, they should receive 1 point less than they would have if they had commented on the possibility of division by zero.