Let's simplify the following fractions:
a) $\frac{3 a^{4} c+2 a^{4} d-3 b^{4} c-2 b^{4} d}{\left[9 c^{2}(a-b)-4 d^{2}(a-b)\right]\left[(a+b)^{2}-2 a b\right]}$,
b) $\frac{a^{2}+b^{2}+c^{2}+2 a b+2 a c+2 b c}{a^{2}-b^{2}-c^{2}-2 b c}$,
c) $\frac{a^{2}-3 a b+a c+2 b^{2}-2 b c}{a^{2}-b^{2}+2 b c-c^{2}}$.
a) Our expression can be written as:

$$
\begin{gathered}
\frac{a^{4}(3 c+2 d)-b^{4}(3 c+2 d)}{(a-b)\left(9 c^{2}-4 d^{2}\right)\left(a^{2}+b^{2}\right)}=\frac{(3 c+2 d)\left(a^{4}-b^{4}\right)}{(a-b)\left(a^{2}+b^{2}\right)\left(9 c^{2}-4 d^{2}\right)}= \\
=\frac{(a+b)(a-b)\left(a^{2}+b^{2}\right)(3 c+2 d)}{(a-b)\left(a^{2}+b^{2}\right)(3 c+2 d)(3 c-2 d)}=\frac{(a+b)}{3 c-2 d}
\end{gathered}
$$

b) Our expression can be written as:

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

c) Our expression can be written as:

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

István Kecskeméti (Bp. V., Piarist g. I. o. t.)