10. Given that
$$
\begin{aligned}
\alpha+\beta+\gamma & =14 \\
\alpha^{2}+\beta^{2}+\gamma^{2} & =84 \\
\alpha^{3}+\beta^{3}+\gamma^{3} & =584
\end{aligned}
$$
find $\max \{\alpha, \beta, \gamma\}$.
10. Answer. 8
Solution. We consider the polynomial
$$
P(t)=t^{3}+b t^{2}+c t+d .
$$

Suppose the root of the equation $P(t)=0$ are $x, y$ and $z$. Then
$$
\begin{array}{c}
-b=x+y+z=14, \\
c=x y+x z+y z=\frac{1}{2}\left((x+y+z)^{2}-x^{2}-y^{2}-z^{2}\right)=\frac{1}{2}\left(14^{2}-84\right)=56
\end{array}
$$
and
$$
x^{3}+y^{3}+z^{3}+3 d=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-x z-y z\right) .
$$

Solving for $b, c$ and $d$, we get $b=-14, c=30$ and $d=-64$. Finally, since $t^{3}-14 t^{2}+30 t-$ $64=0$ implies $t=2$ or $t=4$ or $t=8$, we conclude that $\max \{\alpha, \beta, \gamma\}=8$.