1. Calculate the value of the expression

$$
A=\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-a)(b-c)}+\frac{1}{c(c-a)(c-b)}
$$

if $abc=1$ and $a \neq b \neq c \neq a$.
Solution. We have

$$
\begin{aligned}
A & =\frac{b c(b-c)-a c(a-c)+a b(a-b)}{a b c(a-b)(a-c)(b-c)}=\frac{b^{2} c-b c^{2}-a^{2} c+a c^{2}+a^{2} b-a b^{2}}{a b c(a-b)(a-c)(b-c)} \\
& =\frac{b^{2}(c-a)-b\left(c^{2}-a^{2}\right)+a c(c-a)}{a b c(a-b)(a-c)(b-c)}=\frac{(c-a)\left[b^{2}-b(c+a)+a c\right]}{a b c(a-b)(a-c)(b-c)} \\
& =\frac{(c-a)\left(b^{2}-b c-b a+a c\right)}{a b c(a-b)(a-c)(b-c)}=\frac{(c-a)[b(b-c)-a(b-c)]}{a b c(a-b)(a-c)(b-c)} \\
& =\frac{(c-a)(b-c)(b-a)}{a b c(a-b)(a-c)(b-c)}=\frac{1}{a b c}=1
\end{aligned}
$$