Simplify the expression

$$
\frac{\left(x^{2}-y^{2}\right)^{3}+\left(y^{2}-z^{2}\right)^{3}+\left(z^{2}-x^{2}\right)^{3}}{(x-y)^{3}+(y-z)^{3}+(z-x)^{3}}
$$
I. solution: Based on the identity $\operatorname{Az} u^{3}+v^{3}=(u+v)\left(u^{2}-u v+v^{2}\right)$, we factor the denominator

$$
\begin{aligned}
& (x-y)^{3}+(y-z)^{3}+(z-x)^{3}= \\
= & (x-y+y-z)\left[(x-y)^{2}-(x-y)(y-z)+(y-z)^{2}\right]+(z-x)^{3}= \\
= & (x-z)\left[(x-y)(x-y-y+z)+(y-z)^{2}-(z-x)^{2}\right]= \\
= & (x-z)[(x-y)(x-2 y+z)+(y-z+z-x)(y-z-z+x)]= \\
= & (x-z)[(x-y)(x-2 y+z)-(x-y)(y-2 z+x)]= \\
= & (x-z)(x-y)(x-2 y+z-y+2 z-x)= \\
= & (x-z)(x-y)(3 z-3 y)=3(x-y)(y-x)(z-x) .
\end{aligned}
$$

Similarly, for the numerator

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

Thus, the given fraction is

$$
\frac{3\left(x^{2}-y^{2}\right)\left(y^{2}-x^{2}\right)\left(z^{2}-x^{2}\right)}{3(x-y)(y-z)(z-x)}=(x+y)(y+z)(z+x)
$$

II. solution: Consider the denominator as a polynomial in $x$. Even without actually performing the polynomial transformation, it is easy to see that we get a second-degree polynomial, because the sum of the $x^{3}$ terms from the first and third expressions is 0. The roots of this polynomial are $x=y$ and $x=z$, which can be seen immediately by substitution. Thus, the factorized form of the denominator is the same as the product $(x-y)(x-z)$ and the product of the coefficient of the $x^{2}$ term. The coefficient of $x^{2}$ from the first and third expressions is $-3 y+3 z=3(z-y)$; therefore, the denominator is identical to

$$
3(x-y)(x-z)(z-y)
$$

From this, we get the numerator by substituting $x, y, z$ with $x^{2}, y^{2}, z^{2}$, so the desired fraction is:

$$
\frac{3\left(x^{2}-y^{2}\right)\left(x^{2}-z^{2}\right)\left(z^{2}-y^{2}\right)}{3(x-y)(x-z)(z-y)}=(x+y)(y+z)(z+x)
$$

III. solution: The numerator and the denominator are the sum of the cubes of three numbers whose sum is 0. If $a+b+c=0$, then it can be shown that

$$
a^{3}+b^{3}+c^{3}=3 a b c
$$

Indeed, if

$$
a+b=-c
$$

then cubing both sides,

$$
a^{3}+3 a^{2} b+3 a b^{2}+b^{2}=-c^{3}
$$

or

$$
a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c
$$

This lemma can also be directly derived from the following identity:

$$
a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)
$$

Based on this, our expression can be written as:

$$
\frac{3\left(x^{2}-y^{2}\right)\left(y^{2}-z^{2}\right)\left(z^{2}-x^{2}\right)}{3(x-y)(y-z)(z-x)}=(x+y)(y+z)(z+x) .
$$