Given the function

$$
f(x)=\frac{(x+a)^{2}}{(a-b)(a-c)}+\frac{(x+b)^{2}}{(b-a)(b-c)}+\frac{(x+c)^{2}}{(c-a)(c-b)}
$$

where $a, b$, and $c$ are distinct real numbers. Determine the range of the function.
I. solution. After bringing to a common denominator $(a \neq b, a \neq c, b \neq c)$:

$$
\begin{aligned}
f(x)= & \frac{(b-c)(x+a)^{2}-(a-c)(x+b)^{2}+(a-b)(x+c)^{2}}{(a-b)(a-c)(b-c)}= \\
= & \frac{b x^{2}+2 a b x+a^{2} b-c x^{2}-2 a c x-a^{2} c-a x^{2}-2 a b x-a b^{2}}{(a-b)(a-c)(b-c)}+ \\
& +\frac{c x^{2}+2 b c x+b^{2} c+a x^{2}+2 a c x+a c^{2}-b x^{2}-2 b c x-b c^{2}}{(a-b)(a-c)(b-c)}= \\
= & \frac{a^{2} b-a^{2} c-a b^{2}+b^{2} c+a c^{2}-b c^{2}}{(a-b)(a-c)(b-c)}=\frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)}=1
\end{aligned}
$$

Thus, the range of the function $f$ is: $\{1\}$.

II. solution. Due to the quadratic expressions in the function assignment, $f(x)$ can be at most quadratic $(a \neq b, a \neq c, b \neq c)$. Let's examine the values of $f(x)$ at three different points, specifically at $x=-a$, $x=-b$, and $x=-c$:

$$
\begin{aligned}
f(-a) & =0+\frac{(b-a)^{2}}{(b-a)(b-c)}+\frac{(c-a)^{2}}{(c-a)(c-b)}= \\
& =\frac{b-a}{b-c}+\frac{c-a}{c-b}=\frac{b-a}{b-c}+\frac{a-c}{b-c}=\frac{b-c}{b-c}=1 \\
f(-b) & =\frac{(a-b)^{2}}{(a-b)(a-c)}+0+\frac{(c-b)^{2}}{(c-a)(c-b)}= \\
& =\frac{a-b}{a-c}+\frac{c-b}{c-a}=\frac{a-b}{a-c}+\frac{b-c}{a-c}=\frac{a-c}{a-c}=1 \\
f(-c) & =\frac{(a-c)^{2}}{(a-b)(a-c)}+\frac{(b-c)^{2}}{(b-a)(b-c)}+0= \\
& =\frac{a-c}{a-b}+\frac{b-c}{b-a}=\frac{a-c}{a-b}+\frac{c-b}{a-b}=\frac{a-b}{a-b}=1
\end{aligned}
$$

We found that the function $f(x)$ takes the same value at three different points. Quadratic functions can take the same value at most at two points because quadratic equations have at most two solutions. Linear functions can take the same value at only one point because linear equations have at most one solution.

Thus, our function $f(x)$ cannot be quadratic, it cannot be linear, so it must be a constant function, and its range consists of a single number, 1: $R_{f}=\{1\}$.