1. Find all triples of real numbers $(x, y, z)$ that satisfy the system of equations:

$$
\left\{\begin{array}{l}
x^{3} y^{3} z^{3}=1 \\
x y^{5} z^{3}=2 \\
x y^{3} z^{5}=3
\end{array}\right.
$$
Answer. $\left(\frac{1}{\sqrt[6]{6}}, \frac{\sqrt{2}}{\sqrt[6]{6}}, \frac{\sqrt{3}}{\sqrt[6]{6}}\right) ;\left(-\frac{1}{\sqrt[6]{6}},-\frac{\sqrt{2}}{\sqrt[6]{6}}, \frac{\sqrt{3}}{\sqrt[6]{6}}\right) ;\left(-\frac{1}{\sqrt[6]{6}}, \frac{\sqrt{2}}{\sqrt[6]{6}},-\frac{\sqrt{3}}{\sqrt[6]{6}}\right) ;\left(\frac{1}{\sqrt[6]{6}},-\frac{\sqrt{2}}{\sqrt[6]{6}},-\frac{\sqrt{3}}{\sqrt[6]{6}}\right)$.

Solution. Consider any of the three equations. All three variables in it are in odd powers. This means that either all three numbers $x, y, z$ are positive, or two of them are negative and one is positive. Moreover, if we change the signs of both negative numbers, we will get a positive solution.

Therefore, it is sufficient to find all positive solutions. By changing the signs of these solutions, we will obtain all the others.

So, let's assume that $x, y, z > 0$. Dividing the second equation by the first, we get $y^{2}=2 x^{2}$, i.e., $y=\sqrt{2} x$. Dividing the third equation by the first, we get $z^{2}=3 x^{2}$, i.e., $z=\sqrt{3} x$. Substituting everything into the first equation, we get

$$
6 \sqrt{6} x^{9}=1 \Rightarrow x=\frac{1}{\sqrt[6]{6}}, \quad y=\frac{\sqrt{2}}{\sqrt[6]{6}}, \quad z=\frac{\sqrt{3}}{\sqrt[6]{6}}
$$

Thus, the only positive solution is found. Taking into account the remark about the signs, we get three more solutions, i.e., a total of 4 solutions, as indicated in the answer.

Criteria.

$(-)$ no solution

$(-$.$) derivation of one of the above relations$

$(-/+)$ derivation of relations $x^{2}=\left(y^{2}\right) / 2=\left(z^{2}\right) / 3$

(+/2) solution with sign errors leading to an incorrect answer

$(+/-)$ arithmetic errors in the solution

(+.) solution with incorrect statements about signs, but the final answer is correct

$(+)$ clean solution