Transform the following expression into a product:

$$
(a-b)(b+c+d)(c+a+d)+(b-c)(c+a+d)(a+b+d)+(c-a)(a+b+d)(b+c+d)
$$
I. solution. Let us temporarily introduce the following notations:

$$
b+c+d=A, \quad c+a+d=B, \quad a+b+d=C
$$

With these,

$$
\begin{aligned}
K & =(B-A) A B+(C-B) B C+(A-C) C A= \\
& =B\left(A B-A^{2}+C^{2}-B C\right)+(A-C) C A= \\
& =B\{(A-C)(B-A-C)\}+(A-C) C A= \\
& =(A-C)\left\{B^{2}-A B-B C+C A\right\}=(A-C)(B-A)(B-C)
\end{aligned}
$$

Finally, returning to the original variables,

$$
K=(c-a)(a-b)(c-b)=-(a-b)(b-c)(c-a)
$$

We see that the expression does not depend on $d$.

II. solution. The first term of $K$ (a product of 3 factors) can be partially expanded as follows:

$$
\begin{gathered}
(a-b)\left\{a b+(a+b)(c+d)+(c+d)^{2}\right\}= \\
=a b(a-b)+\left(a^{2}-b^{2}\right)(c+d)+(a-b)(c+d)^{2}
\end{gathered}
$$

And since the second term of $K$ is derived from the first by replacing each $a, b, c, d$ with $b, c, a, d$ respectively, and the third term is derived from the second in the same way, the other two products are:

$$
\begin{aligned}
& b c(b-c)+\left(b^{2}-c^{2}\right)(a+d)+(b-c)(a+d)^{2} \\
& c a(c-a)+\left(c^{2}-a^{2}\right)(b+d)+(c-a)(b+d)^{2}
\end{aligned}
$$

It is easy to see that upon further expansion, the coefficient of $d^{2}$ in the third column is 0, as is the coefficient of $d$ (in the second and third columns separately), and furthermore, that the part of the first and second columns that does not contain $d$ is also 0. Therefore, only the third column remains, which can be further transformed as follows:

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

In the last transformation, we used the insight that the expression in the \{\} can be considered as the left side of a quadratic equation in $c$ reduced to 0, whose roots are $a$ and $b$.

III. solution. Our expression is a polynomial of at most second degree in any of its letters considered as a variable. Considering it as a polynomial in $a$, it can be written in the form

$$
K=k\left(a-a_{1}\right)\left(a-a_{2}\right)
$$

where $k$ is the coefficient of $a^{2}$, and $a_{1}$ and $a_{2}$ are the roots of the polynomial.

It is easy to show that substituting $b$ or $c$ for $a$ results in $K=0$ (for example, when $a=b$, the first term of $K$ is 0, and the other two differ only in the first factors, whose sum is 0), so $a_{1}=b, a_{2}=c$. Furthermore, the coefficient of $a^{2}$ is

$$
k=(b+c+d)+(b-c)-(b+c+d)=b-c
$$

thus

$$
K=(b-c)(a-b)(a-c)
$$

Jenö Reiczigel (Budapest, Fazekas M. Gymnasium, 2nd year)