Solve the following system of equations:

$$
\begin{gathered}
x^{4}\left(y^{2}-z^{2}\right)=243 \\
y^{4}\left(z^{2}-x^{2}\right)=-128 \\
z^{4}\left(x^{2}-y^{2}\right)=5
\end{gathered}
$$
Given $x^{2}=a, y^{2}=b$ and $z^{2}=c$, our system of equations can also be written as:

$$
\begin{array}{rlrl}
\text { (I) } \quad b-c=\frac{243}{a^{2}}, & \text { or } & \text { (II) } \quad a b-a c=\frac{243}{a}, \\
c-a=-\frac{128}{b^{2}}, & b c-a b=-\frac{128}{b} \\
a-b=\frac{5}{c^{2}}, & a c-b c=\frac{5}{c} .
\end{array}
$$

From $(I)$,

$$
243 b^{2} c^{2}-128 a^{2} c^{2}+5 a^{2} b^{2}=0
$$

From $(II)$,

$$
\frac{243}{a}-\frac{128}{b}+\frac{5}{c}=0
$$

$$
243 b c-128 a c+5 a b=0
$$

From $(I)$ and $(II)$,

$$
\begin{gathered}
a_{1}=9 c ; \quad a_{2}=-\frac{201}{41} c \\
b_{1}=4 c ; \quad b_{2}=-\frac{12864}{4579} c
\end{gathered}
$$

Substituting the values of $a_{1}, b_{1}$ and $a_{2}, b_{2}$ into the third equation of the system under $(I)$, we obtain two cubic equations in $c$. If we substitute the values of $a_{1}$ and $b_{1}$, we get

$$
c^{3}=1
$$

from which the real value of $c$ is

$$
c=1
$$

and thus

$$
z=\sqrt{c}= \pm 1, \quad y=\sqrt{b}= \pm 2 \sqrt{c}= \pm 2 \quad \text { and } \quad x=\sqrt{a}= \pm 3
$$

(Balázs Tóth, Eger.)

The problem was also solved by: Ádámffy E., Dömény J., Fekete M., Haar A., Harsányi Z., Kürti I., Rássy P., the V. district gymnasium mathematics circle.