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 $\alpha=7$, $xyz=-1$, $xy+xz+yz=1$, and 
$$ \begin{aligned} x^{5}+y^{5}+z^{5}= & \left(x^{4}+y^{4}+z^{4}\right)(x+y+z) \\ & -\left(x^{3}+y^{3}+z^{3}\right)(x y+x z+y z)+x y z\left(x^{2}+y^{2}+z^{2}\right) \\ = & 35 \cdot 3-15 \cdot 1+7 \cdot(-1)=83 . \end{aligned} $$