Solve the following system of equations:

$$
\begin{array}{ll}
x^{3} y^{2} z=2 & z^{3} u^{2} x=32 \\
y^{3} z^{2} u=8 & u^{3} x^{2} y=8
\end{array}
$$
It is clear that none of the unknowns can be 0, on the other hand, since the exponent of $y$ on the left side of equation (1) is even, while the exponents of $x$ and $z$ are odd, $x$ and $z$ must have the same sign (this can also be read from (3)). Similarly, it follows from (2) (or (4)) that $y$ and $u$ also have the same sign.

We can take advantage of the similar structure of the equations by multiplying them, so that the exponent of each unknown will be the same.

$$
x^{6} y^{6} z^{6} u^{6}=2 \cdot 8 \cdot 32 \cdot 8=2^{12}
$$

Taking the sixth root of this, since the product of the unknowns is positive according to the sign determinations, we get

$$
x y z u=2^{2}=4
$$

Dividing (2) by (1)

$$
\frac{y z u}{x^{3}}=4
$$

and dividing this by (5)

$$
x^{4}=1, \quad \text { from which } \quad x= \pm 1
$$

Similarly, the other unknowns can be calculated: dividing (5) by the quotient of (3) and (2) gives $y= \pm 1$, then dividing (4) and (3), and finally dividing the quotient of (1) and (4) gives $z= \pm 2$, and $u= \pm 2$.

From the above, we also see that the signs of the $x, z$ and $y, u$ pairs of unknowns are independent of each other, so the following four value systems can provide a solution:

$$
\begin{array}{llll}
x_{1}=1, & z_{1}=2, & y_{1}=1, & u_{1}=2 \\
x_{2}=1, & z_{2}=2, & y_{2}=-1, & u_{2}=-2 \\
x_{3}=-1, & z_{3}=-2, & y_{3}=1, & u_{3}=2 \\
x_{4}=-1, & z_{4}=-2, & y_{4}=-1, & u_{4}=-2 .
\end{array}
$$

The check shows that each of them satisfies the system of equations. This solves our problem.

Remarks. 1. It is equally simple to eliminate if we divide the product of (2) and (3) by the product of (1) and (4), and the product of (3) and (4) by the product of (1) and (2)

$$
\frac{x y^{3} z^{5} u^{3}}{x^{5} y^{3} z u^{3}}=\left(\frac{z}{x}\right)^{4}=\frac{8 \cdot 32}{2 \cdot 8}=16, \quad \text { or } \quad\left(\frac{u}{y}\right)^{4}=16
$$

from which - based on the observation about the signs -

$$
z=2 x, \quad u=2 y
$$

Thus, from (1) and (2)

$$
x^{4} y^{2}=x^{2} y^{4}=1, \quad\left(\frac{x}{y}\right)^{2}=1, \quad y= \pm x
$$

2. It is natural to take the logarithm of both sides of equations (1) - (4), thus we get a system of linear equations for the logarithms of the unknowns. Since the powers of 2 are on the right side, it is appropriate to write the equality of the logarithms with base 2.

$$
\begin{gathered}
{ }^{2} \log x=X, \quad{ }^{2} \log y=Y, \quad{ }^{2} \log z=Z, \quad{ }^{2} \log u=U \text { with } \\
3 X+2 Y+Z=1 \\
3 Y+2 Z+U=3 \\
3 Z+2 U+X=5 \\
3 U+2 X+Y=3
\end{gathered}
$$

We do not get essentially different solutions this way either, only addition and subtraction instead of multiplication and division, and multiplication and division instead of exponentiation and root extraction. On the other hand, this way we can only get positive solutions.

Many contestants tried this solution, but they used logarithms with base 10 and replaced the right-hand sides with their approximate values to four decimal places.