2. Given various real numbers $a, b$ and $c$. The quadratic trinomial $f(x)$ satisfies the relations $f(a)=b c, f(b)=c a$ and $f(c)=a b$. Find $f(a+b+c)$.
Answer: $a b+b c+c a$.

Solution. Let $f(x)=p x^{2}+q x+r$. Then from the equalities

$$
b c=f(a)=p a^{2}+q a+r \quad \text { and } \quad c a=f(b)=p b^{2}+q b+r
$$

it follows that $c(b-a)=p\left(a^{2}-b^{2}\right)+q(a-b)$, hence $c=-p(a+b)-q$, since $a \neq b$. Similarly, we obtain the relation $b=-p(a+c)-q$. Subtract the first equality from the second:

$$
b-c=p(a+b)-(a+c))=p(b-c)
$$

Therefore, $p=1$. Then $q=-(a+b+c)$ and $r=f(a)-p a^{2}-q a=a b+b c+c a$. Thus,

$$
\begin{aligned}
f(a+b+c) & =p(a+b+c)^{2}+q(a+b+c)+r= \\
& =(a+b+c)^{2}-(a+b+c)^{2}+(a b+b c+c a)=a b+b c+c a
\end{aligned}
$$