Simplify the following expression:

$$
\frac{(a+b)\left(a^{3}+b^{3}\right)}{(b-c)(c-a)}-\frac{(b+c)\left(b^{3}+c^{3}\right)}{(a-c)(a-b)}+\frac{(a+c)\left(a^{3}+c^{3}\right)}{(b-c)(a-b)}
$$
I. solution. First, let's form the sum of the first two fractions of $K$. Taking the common denominator as the product $N=(a-b)(b-c)(c-a)$, the numerator can be transformed as follows:

$$
\begin{aligned}
S & =\left(a^{2}-b^{2}\right)\left(a^{3}+b^{3}\right)+\left(b^{2}-c^{2}\right)\left(b^{3}+c^{3}\right)= \\
& -\left(c^{5}-a^{5}\right)+\left(c^{3}-a^{3}\right) b^{2}-\left(c^{2}-a^{2}\right) b^{3}
\end{aligned}
$$

Here, all three bracketed expressions are divisible by $c-a$, so $S / N$ can be simplified by $c-a$.

After simplification, the denominators of the sum of the first two terms of $S / N$ are the same as that of the third term of $K$, so they can be added directly. The sum of their numerators can be transformed as follows:

$$
\begin{aligned}
& -\left(c^{4}+c^{3} a+c^{2} a^{2}+c a^{3}+a^{4}\right)+\left(c^{2}+c a+a^{2}\right) b^{2}-(c+a) b^{3}+ \\
& +(a+c)\left(a^{3}+c^{3}\right)=b^{2} c(a-b)+a b^{2}(a-b)-c^{2}\left(a^{2}-b^{2}\right)= \\
& =(a-b)\left[b^{2} c+a b^{2}-c^{2}(a+b)\right]=(a-b)\left[b c(b-c)+a\left(b^{2}-c^{2}\right)\right]= \\
& =(a-b)(b-c)(b c+a b+a c)
\end{aligned}
$$

Dividing this by the common denominator, the expression can be written as

$$
K=a b+b c+c a
$$

assuming that $a, b, c$ do not have two equal values.

Bálint Török (Budapest, Fazekas M. Gyak. G. II. o. t.)

II. solution. Bringing $K$ to a common denominator, it can be written in the form $S_{1} / N_{1}$, where

$S_{1}=\left(a^{2}-b^{2}\right)\left(a^{3}+b^{3}\right)+\left(b^{2}-c^{2}\right)\left(b^{3}+c^{3}\right)+\left(c^{2}-a^{2}\right)\left(a^{3}+c^{3}\right)$, $N_{1}=(a-b)(b-c)(c-a)$

In $S_{1}$, substituting $a$ for $b$ yields identically 0, so the numerator has a factor of $a-b$. Similarly, we find that $S_{1}$ is divisible by $b-c$ and $c-a$ as well. Thus, it is expected that the product of these three factors, i.e., $N_{1}$, can also be factored out of $S_{1}$, so $S_{1}=N_{1} K_{1}$, where $K_{1}$ is a symmetric polynomial, since $K$ is also symmetric in the variables.

Since only fifth-degree terms appear in $S_{1}$, only second-degree terms can appear in $K_{1}$,

$$
K_{1}=\lambda\left(a^{2}+b^{2}+c^{2}\right)+\mu(a b+b c+c a)
$$

The coefficient of $a^{4} b$ in $S_{1}$ is $0$, and in $N_{1} K_{1}$ it is $-\lambda$, so $\lambda=0$; the coefficient of $a^{3} b^{2}$ in $S_{1}$ is $-1$, and in $N_{1} K_{1}$ it is $-\mu$, so $\mu=1$, thus $K_{1}=a b+b c+c a$.

By multiplication, it can be easily verified that indeed $N_{1} K_{1}=S_{1}$, so $K=K_{1}$.