Simplify the following fraction as much as possible:

$$
\frac{\left(x^{2}-y^{2}\right)^{3}+\left(y^{2}-z^{2}\right)^{3}+\left(z^{2}-t^{2}\right)^{3}+\left(t^{2}-x^{2}\right)^{3}}{(x-y)^{3}+(y-z)^{3}+(z-t)^{3}+(t-x)^{3}}
$$
I. solution: Let us denote our fraction by $N$. We break down the cubes and subtract:

$$
N=\frac{x^{2} y^{4}-x^{4} y^{2}-y^{4} z^{2}+y^{2} z^{4}-z^{4} t^{2}+z^{2} t^{4}-t^{4} x^{2}+t^{2} x^{4}}{x y^{2}-x^{2} y-y^{2} z+y z^{2}-z^{2} t+z t^{2}-t^{2} x+t x^{2}}
$$

We rearrange according to the powers of $y$ and $t$:

$$
\begin{gathered}
N=\frac{y^{4}\left(x^{2}-z^{2}\right)+y^{2}\left(z^{4}-x^{4}\right)+t^{4}\left(z^{2}-x^{2}\right)+t^{2}\left(x^{4}-z^{4}\right)}{y^{2}(x-z)+y\left(z^{2}-x^{2}\right)+t^{2}(z-x)+t\left(x^{2}-z^{2}\right)}= \\
=\frac{\left(z^{2}-x^{2}\right)\left[\left(z^{2}+x^{2}\right)\left(y^{2}-t^{2}\right)+\left(t^{4}-y^{4}\right)\right]}{(z-x)\left[(z+x)(y-t)+\left(t^{2}-y^{2}\right)\right]}=\frac{\left(z^{2}-x^{2}\right)\left(y^{2}-t^{2}\right)\left(z^{2}+x^{2}-t^{2}-y^{2}\right)}{(z-x)(y-t)(z+x-y-t)}
\end{gathered}
$$

Assuming that $x \neq z$ and $y \neq t$, we divide the numerator and the denominator by $(z-x)(y-t)$ and get:

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

Csaba Máthé (Győr, Révai g. I. o. t.)

II. solution: The numerator and the denominator are the sum of the cubes of four numbers whose sum is zero. If $a+b+c+d=0$, then from the equation $a+b=-(c+d)$, cubing both sides we get:

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

which simplifies to:

$$
\begin{gathered}
a^{3}+b^{3}+c^{3}+d^{3}=-3 a b(a+b)-3 c d(c+d)= \\
=-3 a b(a+b)-3 c d[-(a+b)]=3(a+b)(c d-a b)
\end{gathered}
$$

Based on this, the given fraction can be written as:

$$
\begin{gathered}
\frac{3\left(x^{2}-z^{2}\right)\left[\left(z^{2}-t^{2}\right)\left(t^{2}-x^{2}\right)-\left(x^{2}-y^{2}\right)\left(y^{2}-z^{2}\right)\right]}{3(x-z)[(z-t)(t-x)-(x-y)(y-z)]}= \\
=\frac{(x+z)\left[\left(y^{4}-t^{4}\right)-x^{2}\left(y^{2}-t^{2}\right)-z^{2}\left(y^{2}-t^{2}\right)\right]}{\left(y^{2}-t^{2}\right)-x(y-t)-z(y-t)}= \\
=\frac{(x+z)\left(y^{2}-t^{2}\right)\left(y^{2}+t^{2}-x^{2}-z^{2}\right)}{(y-t)(y+t-x-z)}=\frac{(x+z)(y+t)\left(x^{2}-y^{2}+z^{2}-t^{2}\right)}{x-y+z-t}
\end{gathered}
$$

again assuming that $x-z \neq 0, y-t \neq 0$.

Péter Tatai (Bp. XIV., I. István g. II. o. t.)

III. solution: The denominator can be considered as a quadratic polynomial in $x$, since the sum of the cubic terms from the first and fourth terms is zero. Without actually performing the polynomial transformation, it can be directly verified by substitution that the roots of this polynomial are $x=z$ and $x=y-z+t$. The coefficient of $x^{2}$ from the first and fourth terms is: $-3 y+3 t=-3(y-t)$. Therefore, the denominator is identical to the following product:

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

If we substitute $x, y, t, z$ with $x^{2}, y^{2}, t^{2}, z^{2}$ respectively, we get the numerator. Thus, the value of our fraction (if $x \neq z, y \neq t$) is:

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

Ervin Gergely (Bp. IV., Könyves Kálmán g. IV. o. t.)