3-ча 1. Solve the system of equations:

$$
\left\{\begin{aligned}
x+y & =a \\
x^{5}+y^{5} & =b^{5}
\end{aligned}\right.
$$
Let $x y=t$. It is not hard to verify that

$$
x^{5}+y^{5}=(x+y)^{5}-5(x+y)^{3} x y+5(x+y) x^{2} y^{2}=a^{5}-5 a^{3} t+5 a t^{2}
$$

For $t$, we obtain the quadratic equation $t^{2}-a^{2} t+\frac{a^{5}-b^{5}}{5 a}=0$. Solving it, we find $t=\frac{1}{2}\left(a^{2} \pm \sqrt{\frac{a^{5}+4 b^{5}}{5 a}}\right)$. As a result, we obtain the system of equations

$$
\left\{\begin{aligned}
x+y & =a \\
x y & =t
\end{aligned}\right.
$$

Solving this system also reduces to solving a quadratic equation.