3. Determine all triples of natural numbers $(x, y, z)$ such that

$$
2 x^{2} y^{2}+2 y^{2} z^{2}+2 x^{2} z^{2}-x^{4}-y^{4}-z^{4}=576
$$
Solution. We have

$$
\begin{aligned}
4 x^{2} y^{2}-\left(x^{4}+y^{4}+z^{4}+2 x^{2} y^{2}-\right. & \left.2 y^{2} z^{2}-2 x^{2} z^{2}\right)=4 x^{2} y^{2}-\left(x^{2}+y^{2}-z^{2}\right)^{2} \\
& =\left(2 x y-x^{2}-y^{2}+z^{2}\right)\left(2 x y+x^{2}+y^{2}-z^{2}\right) \\
& =\left(z^{2}-(x-y)^{2}\right)\left((x+y)^{2}-z^{2}\right) \\
& =(z-x+y)(z+x-y)(x+y-z)(x+y+z)=576
\end{aligned}
$$

Since the sum of any two of the numbers

$$
z-x+y, z+x-y, x+y-z, x+y+z
$$

is an even number, they must all have the same parity, and since their product is an even number, they must all be even numbers. Let

$$
z-x+y=2 a, z+x-y=2 b, x+y-z=2 c, x+y+z=2 d
$$

Then $16 a b c d=576$, i.e., $a b c d=36$. Without loss of generality, we can assume that $x \geq y \geq z$, from which it follows that $d>c \geq b \geq a$. Further, the number 36 as a product of four factors, one of which is greater than the other three, can be represented in the following ways:

$$
36=6 \cdot 3 \cdot 2 \cdot 1=12 \cdot 3 \cdot 1 \cdot 1=18 \cdot 2 \cdot 1 \cdot 1=36 \cdot 1 \cdot 1 \cdot 1
$$

From $36=6 \cdot 3 \cdot 2 \cdot 1$ we get $a=1, b=2, c=3, d=6$, from which it follows that

$$
x=b+c=5, y=a+c=4 \text { and } z=a+b=3 \text {. }
$$

It is easy to see that the other representations of the number 36 lead to contradictions. Finally,

$$
(x, y, z) \in\{(3,4,5),(5,3,4),(4,5,3),(5,4,3),(3,5,4),(4,3,5)\}
$$