3.4. $\left\{\begin{array}{l}x+y+z=a, \\ x^{2}+y^{2}+z^{2}=a^{2}, \\ x^{3}+y^{3}+z^{3}=a^{3} .\end{array}\right.$
3.4. A n s w e r: $(0,0, a),(0, a, 0)$ 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 $x y z=(x y+y z+x z)(x+y+z)-(x+y((y+z)(z+x)=0$. 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 obtain the following solutions: $(0,0, a)$, $(0, a, 0)$ and $(a, 0,0)$.