3. (GDR 1) Knowing that the system
$$
\begin{aligned}
x+y+z & =3, \\
x^{3}+y^{3}+z^{3} & =15, \\
x^{4}+y^{4}+z^{4} & =35,
\end{aligned}
$$
has a real solution $x, y, z$ for which $x^{2}+y^{2}+z^{2}<10$, find the value of $x^{5}+y^{5}+z^{5}$ for that solution.
3. Let $x, y, z$ be a solution of the given system with $x^{2}+y^{2}+z^{2}=\alpha<10$. Then
$$
x y+y z+z x=\frac{(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)}{2}=\frac{9-\alpha}{2} .
$$

Furthermore, $3 x y z=x^{3}+y^{3}+z^{3}-(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$, which gives us $x y z=3(9-\alpha) / 2-4$. We now have
$$
\begin{aligned}
35= & x^{4}+y^{4}+z^{4}=\left(x^{3}+y^{3}+z^{3}\right)(x+y+z) \\
& -\left(x^{2}+y^{2}+z^{2}\right)(x y+y z+z x)+x y z(x+y+z) \\
= & 45-\frac{\alpha(9-\alpha)}{2}+\frac{9(9-\alpha)}{2}-12 .
\end{aligned}
$$

The solutions in $\alpha$ are $\alpha=7$ and $\alpha=11$. Therefore $\alpha=7, x y z=-1$, $x y+x z+y z=1$, and
$$
\begin{aligned}
x^{5}+y^{5}+z^{5}= & \left(x^{4}+y^{4}+z^{4}\right)(x+y+z) \\
& -\left(x^{3}+y^{3}+z^{3}\right)(x y+x z+y z)+x y z\left(x^{2}+y^{2}+z^{2}\right) \\
= & 35-3-15 \cdot 1+7 \cdot(-1)=83 .
\end{aligned}
$$