Solve the following system of equations:

$$
x+y+z=0 \\
x^{3}+y^{3}+z^{3}=0 \\
x^{1999}-y^{1999}+z^{1999}=2^{2000} .
$$
The identity $\mathrm{Az}(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$ is applied twice:

$0=(x+y+z)^{3}=((x+y)+z)^{3}=(x+y)^{3}+z^{3}+3(x+y) z(x+y+z)==x^{3}+y^{3}+3 x y(x+y)+z^{3}+3(x+y) z(x+y+z)=$

The solutions to the system of the first two equations are precisely those triples of numbers $x, y, z$ where one element is 0 and the other two are each other's opposites. If $z=0$ and $y=-x$, then according to the third equation $2 x^{1999}=2^{2000}$, from which we get the solution $x=2, y=-2, z=0$. If $y=0$ and $z=-x$, then similarly $0=2^{2000}$ results, which does not provide a solution. Finally, if $x=0$ and $z=-y$, then $-2 y^{1999}=2^{2000}$ according to the third equation, yielding the solution $x=0, y=-2, z=2$.

Jesch David (Nagykanizsa Batthyány L. Gymn., 9th grade) Torda Péter (Gyöngyös, Berze Nagy J. Gymn., 11th grade)