2. Find all triples of real numbers $(x, y, z)$ satisfying the system of equations

$$
\begin{aligned}
x^{3}+y^{3} & =9 z^{3} \\
x^{2} y+y^{2} x & =6 z^{3} .
\end{aligned}
$$
SOLUTION. By adding three times the second equation to the first equation, we obtain the equation

$$
x^{3}+3 x^{2} y+3 x y^{2}+y^{3}=27 z^{3} \text {. }
$$

By rearranging it, we get

$$
(x+y)^{3}=(3 z)^{3}, \quad \text { i.e. } \quad x+y=3 z .
$$

Substituting this expression into the left side of the second equation of the system, we get

$$
x^{2} y+x y^{2}=x y(x+y)=3 x y z, \quad \text { i.e. } \quad 3 x y z=6 z^{3} .
$$

We distinguish two cases.

If $z=0$, the last equation is satisfied for all $x, y \in \mathbb{R}$. From the first equation of the system, we get $x^{3}+y^{3}=0$, i.e., $y=-x$. The solution is any triplet $(t,-t, 0)$, where $t$ is any real number.

If $z \neq 0$, then $x y=2 z^{2}$. Together with the equation $x+y=3 z$, we get the system

$$
\begin{aligned}
x+y & =3 z, \\
x y & =2 z^{2}
\end{aligned}
$$

of two equations in two unknowns $x, y$ with parameter $z$. By eliminating, for example, the unknown $y$, we get the quadratic equation

$$
x^{2}-3 z x+2 z^{2}=0 \text {. }
$$

From the relationships between the roots and coefficients of the quadratic equation, we obtain the solution in the form $x=z$, $y=2 z$ or $x=2 z, y=z$. Therefore, the solution is any triplet $(t, 2 t, t)$ and $(2 t, t, t)$, where $t$ is any real number (different from zero).

Conclusion: The system has solutions $(t, 2 t, t)$ and $(2 t, t, t)$ for every $t \neq 0$, and $(t, -t, 0)$ for every $t$, and no other solutions.