## Task B-1.2.

If $\frac{a}{a+b^{2}}+\frac{b}{b+a^{2}}=\frac{4}{5}$ and $a b=-1$, what is $a^{9}+b^{9}$?
## Solution.

By rearranging the equation $\frac{a}{a+b^{2}}+\frac{b}{b+a^{2}}=\frac{4}{5}$, we get

$$
\begin{array}{cc}
\frac{a b+a^{3}+a b+b^{3}}{\left(a+b^{2}\right)\left(b+a^{2}\right)}=\frac{4}{5} & \\
5\left(a^{3}+2 a b+b^{3}\right)=4\left(a b+a^{3}+b^{3}+a^{2} b^{2}\right) & 1 \text { point } \\
a^{3}+b^{3}=4 a^{2} b^{2}-6 a b . & 1 \text { point }
\end{array}
$$

According to the condition of the problem $a b=-1$, so $a^{3}+b^{3}=10$.

Cubing this equality, we get

$$
\begin{array}{cc}
\left(a^{3}+b^{3}\right)^{3}=1000 & \\
a^{9}+3 a^{6} b^{3}+3 a^{3} b^{6}+b^{9}=1000 & 1 \text { point } \\
a^{9}+b^{9}=1000-3 a^{3} b^{3}\left(a^{3}+b^{3}\right) . & 1 \text { point }
\end{array}
$$

Since $a b=-1$ and $a^{3}+b^{3}=10$, it follows that $a^{9}+b^{9}=1000-3(-1)^{3} \cdot 10=1030$.