85. From the solution of the previous problem, we have $a+b=c n ; b+c=a n ; a+c=b n$. After subtracting, we get $a-c=n(c-a)$. The equality is possible when $a-c=0$, i.e., $a=c$. Similarly, $a=b$, so $a=b=c$, and from this, $\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=2$. Therefore, $\frac{(a+b)(b+c)(a+c)}{a b c}=8$.
85. To which number can the expression

$$
\frac{(a+b)(b+c)(a+c)}{a b c}
$$

be equal if the numbers $a, b$ and $c$ satisfy the condition of problem 84?