Express the following as a product of polynomials in $a$, $b$, and $c$:

$$
a(b-c)^{3}+b(c-a)^{3}+c(a-b)^{3}
$$
I. solution: Expanding the given expression $P(a, b, c)$, six terms of the form $\mathrm{a} \pm 3 x^{2} y z$ disappear - here $x, y, z$ represent $a, b, c$ in some order - (readers should verify this by mental calculation without writing down the terms; the Ed.), and the remaining six terms, when arranged according to the powers of $c$, and the equal power differences are factored into products, the factor $(a-b)$ can be factored out:

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

Now, rearranging the terms in the large bracket according to the powers of $a$, and factoring out $(b-c)$:

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

Finally, rearranging the terms in the large bracket according to the powers of $b$, and factoring out $(c-a)$:

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

and with this, we have expressed the expression as the product of four linear polynomials in $a, b$, and $c$, and no further factorization is possible.

Judit Várhelyi (Bp. I., Építőanyagipari technikum IV.o. t)

II. solution: The expression $P(x, b, c)$, as a polynomial in $x$, has zeros at $x=b$ and $x=c$:

$$
\begin{aligned}
P(b, b, c) & =b(b-c)^{3}+b(c-b)^{3}=0 \\
c(c, b, c) & =c(b-c)^{3}+c(c-b)^{3}=0
\end{aligned}
$$

Therefore, substituting $a$ back for $x$, the factors $a-b$ and $a-c$ can be factored out.

Similarly, it follows that the factors $b-a$ and $b-c$, as well as $c-a$ and $c-b$, can also be factored out. However, of the six factors obtained, two differ only in sign, so we have three essentially different factors.

On the other hand, we see that substituting $b, c, a$ for $a, b, c$, and then $c, a, b$, does not change the expression. To ensure that this property remains visible after factoring out the above factors, let the first factor be $a-b$, and the other two be the differences obtained from it by the above substitutions; $b-c$ and $c-a$. The product $Q(a, b, c)=(a-b)(b-c)(c-a)$ of the factors factored out is also a homogeneous expression in $a, b, c$, just like $P(a, b, e)$, with degrees 3 and 4, respectively. Therefore, the division of $P(a, b, c)$ by $Q(a, b, c)$ results in a homogeneous expression $R(a, b, c)$, whose degree is $4-3=1$. Let the coefficients of $a, b, c$ in this be $A, B, C$, respectively, that is, $R(a, b, c)=A a+B b+C c$. The above substitutions must also carry this into itself. From the requirement that

$$
A a+B b+C c=A b+B c+C a=A c+B a+C b
$$

is an identity, that is, it holds for any set of values of $a, b, c$, it follows (e.g., by substituting $a=1, b=c=0$), that $A=B=C$, so $R(a, b, c)=A(a+b+c)$ and

$$
P(a, b, c)=A(a-b)(b-c)(c-a)(a+b+c)
$$

Finally, from the equality of the coefficients of $a^{3}$ on the left and right sides, $-b+c=-A(b-c)$, so $A=1$ and

$$
P(a, b, c)=(a-b)(b-c)(c-a)(a+b+c)
$$

László Flanek (Bp. I., Toldy F. g. III. o. t.)

Remark. $P(x, b, c)$ as a polynomial in $x$ is of the third degree. Finding the third zero would also give us a factor that can be factored out (indeed, this is the consistent continuation of the determination of the first two factors). According to the first arrangement in the I. solution, the coefficient of $x^{2}$ in the expanded form of $P(x, b, c)$ is 0, so the sum of the roots of the equation $P(x, b, c)=0$ is 0, and from $x_{1}=b$ and $x_{2}=c$, we get $x_{3}=-b-c$, the third root factor: $a-x_{3}=a+b+c$. For the purpose of the root factor form, the coefficient of $x^{3}$ is similar to the first arrangement in the I. solution, $c-b$, so $P(a, b, c)=(c-b)(a-b)(a-c)(a+b+c)$, and from this, the above symmetric form is obtained by multiplying by $(-1)^{2}$.

III. solution: To facilitate the transformations, let's try to introduce a single letter for each of the differences whose cube appears, since $(b-c)+(c-a)+(a-b)=0$, so the third can be expressed with two. Let $b-c=m, c-a=n$, then $a-b=-m-n$, and from the first two equalities, $b$ and $a$ can be expressed, and the given expression can be transformed into a polynomial in $c, m$ and $n$: $a=c-n, b=c+m$, and the given expression:

$$
\begin{gathered}
(c-n) m^{3}+(c+m) n^{3}-c(m+n)^{3}=\left[m^{3}+n^{3}-(m+n)^{3}\right] c+m n\left(n^{2}-m^{2}\right)= \\
=-3 m n(m+n) c+m n(m+n)(n-m)=m n(m+n)(-3 c+n-m)
\end{gathered}
$$

finally, returning to the original variables:

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