Problem 5. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}=(y-z)^{2}-3 \\
y^{2}=(z-x)^{2}-9 \\
z^{2}=(x-y)^{2}+27
\end{array}\right.
$$
Answer: $(-1,-4,-6),(1,4,6)$.

Solution. Move the square of the difference in each equation to the left side and apply the formula for the difference of squares:

$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=-3 \\
(y-z+x)(y+z-x)=-9 \\
(z-x+y)(z+x-y)=27
\end{array}\right.
$$

Let $X=-x+y+z, Y=x-y+z, Z=x+y-z$. Then

$$
\left\{\begin{aligned}
Y Z & =-3 \\
Z X & =-9 \\
X Y & =27
\end{aligned}\right.
$$

Multiplying all the obtained equations, we have $(X Y Z)^{2}=3 \cdot 9 \cdot 27$, from which $X Y Z=27$ or $X Y Z=$ -27. Consider the case $X Y Z=27$. In this case, $X=(X Y Z) /(Y Z)=-9, Y=-3, Z=1$; then $x=\frac{Y+Z}{2}=$ $-1, y=-4, z=-6$. The second case is considered similarly and in it $x=1, y=4, z=6$.