Problem 8. Solve the system

$$
\left\{\begin{aligned}
x+y+z & =15 \\
x^{2}+y^{2}+z^{2} & =81 \\
x y+x z & =3 y z
\end{aligned}\right.
$$
Answer: $(6 ; 3 ; 6),(6 ; 6 ; 3)$.

Solution. Let $y+z$ be denoted by $p$, and $yz$ by $q$. Then the system becomes

$$
\left\{\begin{aligned}
x+p & =15 \\
x^{2}+p^{2}-2 q & =81 \\
x p & =3 q .
\end{aligned}\right.
$$

From the last two equations, $(x+p)^{2}-\frac{8}{3} x p=81$, and considering the first equation, $x p=54$. Solving the system

$$
\left\{\begin{aligned}
x+p & =15 \\
x p & =54
\end{aligned}\right.
$$

we find that either $x=6, p=9$, or $x=9, p=6$. The second case is impossible, as the system

$$
\left\{\begin{aligned}
y+z & =6 \\
y z & =18
\end{aligned}\right.
$$

has no solutions, while the first case gives the above answer.