3-ча 1. Solve the system:

$$
\left\{\begin{aligned}
x+y+z & =a \\
x^{2}+y^{2}+z^{2} & =a^{2} \\
x^{3}+y^{3}+z^{3} & =a^{3}
\end{aligned}\right.
$$
Solve 1. Answer: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$.

The identity $(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+x z)$ shows that

$$
x y+y z+x z=0
$$

The identity $(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)$ shows that $(x+y)(y+z)(z+x)=0$. Considering equation (1), we get $3 x y z=0$. If $x=0$, then equation (1) shows that $y z=0$. Therefore, either $y=0$ and $z=a$, or $z=0$ and $y=a$. The other cases are similarly analyzed. In the end, we get the following solutions: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$.