Certainly, here is the translated text with the original formatting preserved:

```
Transform the following expression into a product:

\[
\begin{gathered}
2(a+b)(b+c)(a+3 b+2 c)+2(b+c)(c+a)(b+3 c+2 a)+ \\
\quad+2(c+a)(a+b)(c+3 a+2 b)+9(a+b)(b+c)(c+a) .
\end{gathered}
\]
```
For the sake of simplification, let temporarily:

$$
a+b=x, \quad b+c=y, \quad c+a=z
$$

With partial multiplication and appropriate regrouping, the factor $x+2 y$ can be factored out:

$$
\begin{gathered}
K=2 x y(x+2 y)+2 y z(y+2 z)+2 z x(z+2 x)+9 x y z= \\
=2 x y(x+2 y)+\left(2 y^{2} z+x y z\right)+\left(2 x z^{2}+4 y z^{2}\right)+\left(4 x^{2} z+8 x y z\right)= \\
=2 x y(x+2 y)+y z(x+2 y)+2 z^{2}(x+2 y)+4 x z(x+2 y)= \\
=(x+2 y)\left(y z+2 x y+2 z^{2}+4 x z\right)
\end{gathered}
$$

The second factor can be further factored:

$$
y(z+2 x)+2 z(z+2 x)=(y+2 z)(z+2 x)
$$

Therefore, reverting to the original variables:

$$
\begin{gathered}
K=(x+2 y)(y+2 z)(z+2 x)= \\
=(a+3 b+2 c)(b+3 c+2 a)(c+3 a+2 b)
\end{gathered}
$$

László Köszegi (Baja, III. Béla g. IV. o. t.)

Note: If after introducing the notations $x, y, z$, we arrange the expression according to the powers of $x$,

$$
(2 y+4 z) x^{2}+\left(4 y^{2}+9 y z+2 z^{2}\right) x+\left(2 y^{2} z+4 y z^{2}\right)
$$

is obtained. Here, the first and last terms are divisible by $y+2 z$, and testing shows that the middle term is also. After factoring out, finding the zeros of the remaining expression for $x$ leads us back to the factorization obtained above.

Jenö Lehel (Budapest, Apáczai Csere J. gyak. g. IV. o. t.)