Let's simplify the expression

$$
\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(c-a)(a-b)}+\frac{(c-a)^{2}}{(a-b)(b-c)}
$$
# I. Solution:

$$
\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(c-a)(a-b)}+\frac{(c-a)^{2}}{(a-b)(b-c)}
$$

brought to a common denominator

$$
\frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(a-b)(b-c)(c-a)}
$$

transforming the binomial expression into a polynomial and performing the simplification, the fraction becomes

$$
\begin{aligned}
& \frac{-3 a^{2} b+3 a b^{2}-3 b^{2} c+3 b c^{2}-3 a c^{2}+3 a^{2} c}{a^{2} b+a^{2} c-a c^{2}-b^{2} c+b^{2} a+b^{2} c}= \\
& =3 \frac{-a^{2} b+a b^{2}-b^{2} c+b c^{2}-a c^{2}+a^{2} c}{-a^{2} b+a b^{2}-b^{2} c+b c^{2}-a c^{2}+a^{2} c}=3
\end{aligned}
$$

Pál Csonka (II. class)

II. Solution: After bringing to a common denominator, do not multiply out the denominator and do not expand the parentheses in the numerator, but transform the numerator into a product, as this makes simplification possible.

$$
\begin{gathered}
\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(c-a)(a-b)}+\frac{(c-a)^{2}}{(a-b)(b-c)}= \\
=\frac{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}{(b-c)(c-a)(a-b)}
\end{gathered}
$$

Transform the sum of the first two cubes into a product:

$$
\frac{[(a-b)+(b-c)]\left[(a-b)^{2}-(a-b)(b-c)+(b-c)^{2}\right]+(c-a)^{3}}{(b-c)(c-a)(a-b)}
$$

simplifying and factoring out $(c-a)$

$$
\frac{-(c-a)\left[(a-b)^{2}+(b-c)^{2}-(c-a)^{2}-(a-b)(b-c)\right]}{(b-c)(c-a)(a-b)}
$$

transforming the difference of squares into a product

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

III. Solution: In the previous solution, after factoring out the sum of the bases from the sum of the two cubes, we saw that the numerator is divisible by $(c-a)$. However, without going through the previous calculations, we can immediately conclude that due to symmetry, the numerator is also divisible by $(b-c)$ and $(a-b)$. Since the numerator is a third-degree expression, it follows that it can only differ from the denominator by a constant that is independent of $a, b$, and $c$. The value of this constant can be most easily obtained by substituting suitable numbers for $a, b$, and $c$.

Let $a=1, b=0, c=-1$. Then

$$
\begin{gathered}
a-b=1, \quad b-c=1, \quad c-a=-2 \\
\frac{1^{3}+1^{3}+(-2)^{3}}{1 \cdot 1 \cdot(-2)}=\frac{-6}{-2}=3
\end{gathered}
$$

IV. Solution: Let $a-b$ be $\alpha$, $b-c$ be $\beta$, and $c-a$ be $\gamma$.

Since $\alpha+\beta+\gamma=0, \gamma=-(\alpha+\beta)$.

Substituting these into the original fraction expression

$$
\begin{gathered}
\frac{\alpha^{2}}{\beta \gamma}+\frac{\beta^{2}}{\gamma \alpha}+\frac{\gamma^{2}}{\alpha \beta}=\frac{\alpha^{3}+\beta^{3}+\gamma^{3}}{\alpha \beta \gamma}=\frac{\alpha^{3}+\beta^{3}-(\alpha+\beta)^{3}}{\alpha \beta \gamma}= \\
=\frac{-3 \alpha^{2} \beta-3 \alpha \beta^{2}}{-\alpha \beta(\alpha+\beta)}=\frac{3(-\alpha \beta)(\alpha+\beta)}{-\alpha \beta(\alpha+\beta)}=3
\end{gathered}
$$