Solve the system:

$$
\begin{aligned}
& x+y+z=a \\
& x^{2}+y^{2}+z^{2}=a^{2} \\
& x^{3}+y^{3}+z^{3}=a^{3}
\end{aligned}
$$
From the identity $(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+x z)$, we find that $x y+y z+x z=0$. From the identity $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\right)$ (see problem $\underline{61005}$ g), we obtain that $x y z=0$. Therefore, $x, y, z$ are the roots of the cubic equation $x^{3}-a x^{2}=0$.

## Answer

$\{a, 0,0\}$.