8.4. It is known that $a b c=1$. Calculate the sum

$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}
$$
Solution: Note that

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

Similarly, by replacing 1 with the number $a b c$, we have

$$
\frac{1}{1+c+c a}=\frac{a b}{1+a+a b}=\frac{a b^{2} c}{1+b+b c}=\frac{b}{1+b+b c} .
$$

Then

$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}=\frac{b c}{1+b+b c}+\frac{1}{1+b+b c}+\frac{b}{1+b+b c}=1
$$

Answer: 1.

Recommendations for checking:

| present in the work | points |
| :--- | :--- |
| Correct justified answer | 7 points |
| Correct answer obtained by considering special cases (in any number) | 1 point |
| Correct answer without justification OR incorrect answer OR algebraic transformations not leading to the answer | 0 points |