8.3. The numbers $a, b, c$ satisfy the relation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$. Find $(a+b)(b+c)(a+c)$.
Answer: 0. Solution: Move $\frac{1}{a}$ to the right side, we get $\frac{b+c}{b c}=\frac{-(b+c)}{a(a+b+c)}$. If $b+c \neq 0$, then we will have (multiplying by the denominator)

$$
a^{2}+a b+a c+b c=0 \Leftrightarrow a(a+b)+c(a+b)=0 \Leftrightarrow(a+b)(a+c)=0
$$

Thus, in any case $(a+b)(b+c)(a+c)=0$.