Simplify the following expressions to their simplest form:
a) $\left\{3 x^{2}\left(a^{2}+b^{2}\right)-3 a^{2} b^{2}+3\left[x^{2}+(a+b) x+a b\right] \cdot[x(x-a)-b(x-a)]\right\}: x^{2}$

$$
\text { b) } \frac{1}{a-b}-\frac{3 a b}{a^{3}-b^{3}}-\frac{b-a}{a^{2}+a b+b^{2}}
$$

c) $\frac{a}{(a-2 b)(a-c)}+\frac{2 b}{(2 b-c)(2 b-a)}+\frac{c}{(c-a)(c-2 b)}$
a) The expression in the square brackets can be written as:

$$
\begin{aligned}
& x^{2}+(a+b) x+a b=(x+a)(x+b) \\
& \quad x(x-a)-b(x-a)=(x-a)(x-b)
\end{aligned}
$$

The triple product of these two expressions is

$$
3\left(x^{2}-a^{2}\right)\left(x^{2}-b^{2}\right)=3 x^{4}+3 a^{2} x^{2}-3 b^{2} x^{2}+3 a^{2} b^{2}
$$

and thus the dividend in the curly brackets is

$$
3 a^{2} x^{2}+3 b^{2} x^{2}-3 a^{2} b^{2}+3 x^{4}-3 a^{2} x^{2}-3 b^{2} x^{2}+3 a^{2} b^{2}=3 x^{4}
$$

Therefore, the desired quotient is

$$
3 x^{4}: x^{2}=3 x^{2}
$$

b) Since $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$ and $-(b-a)=a-b$, we have

$$
\begin{gathered}
\frac{1}{a-b}-\frac{3 a b}{a^{3}-b^{3}}-\frac{b-a}{a^{3}+b+b^{2}}=\frac{a^{2}+a b+b^{2}-3 a b+(a-b)^{2}}{(a-b)\left(a^{2}+a b+b^{2}\right)}= \\
=\frac{2 a^{2}-4 a b+2 b^{2}}{(a-b)\left(a^{2}+a b+b^{2}\right)}=\frac{2(a-b)^{2}}{(a-b)\left(a^{3}+a b+b^{2}\right)}=\frac{2(a-b)}{a^{2}+a b+b^{2}} \\
c) \quad \frac{a}{(a-2 b)(a-c)}+\frac{2 b}{(2 b-c)(2 b-a)}+\frac{c}{(c-a)(c-2 b)}= \\
=\frac{a}{(a-2 b)(a-c)}-\frac{2 b}{(a-2 b)(2 b-c)}+\frac{c}{(2 b-c)(a-c)}
\end{gathered}
$$

It can be shown that for any two terms of this sum, they are equal to the third term with the opposite sign, provided that $a \neq c, a \neq 2 b, c \neq 2 b$. For example, the first two terms:

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

and thus our expression is identically equal to 0.

Denes Redly (Pannonhalma g. II. o. t.)