3. For the quadratic trinomial $f(x)=a x^{2}+b x+c$, it is known that

$$
f\left(\frac{a-b-c}{2 a}\right)=f\left(\frac{c-a-b}{2 a}\right)=0
$$

Find the value of the product $f(-1) \cdot f(1)$.
Answer: 0.

## Solution:

$$
f\left(\frac{a-b-c}{2 a}\right)=\frac{a(a-b-c)^{2}}{4 a^{2}}+\frac{b(a-b-c)}{2 a}+c=\frac{(a-b+c)(a+b+c)}{4 a}=\frac{f(-1) \cdot f(1)}{4 a}=0
$$