SUBIECTUL II: Let $a, b, c$ be non-zero rational numbers, such that any two of them are different. Knowing that $c=a+b$, calculate: $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)$.
## SUBJECT II

$$
\begin{gathered}
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)= \\
=\left(\frac{a-b}{a+b}+\frac{b-a-b}{a}+\frac{a+b-a}{b}\right) \cdot\left(\frac{a+b}{a-b}+\frac{b}{b-a-b}+\frac{b}{a+b-a}\right)= \\
=\left(\frac{a-b}{a+b}-1+1\right)\left(\frac{a+b}{a-b}-1+1\right)=\frac{a-b}{a+b} \cdot \frac{a+b}{a-b}=1
\end{gathered}
$$