1. Given that $f(w, w, w) = 0$ for all $w \in \mathbb{R}$, we need to show that there exist polynomials $A(x, y, z)$, $B(x, y, z)$, and $C(x, y, z)$ such that $A + B + C = 0$ and
$$f(x, y, z) = A(x, y, z) \cdot (x - y) + B(x, y, z) \cdot (y - z) + C(x, y, z) \cdot (z - x).$$

2. Let $C = -A - B$. Then we can rewrite $f$ as:
$$f = A \cdot (x - y) + B \cdot (y - z) + (-A - B) \cdot (z - x).$$
Simplifying this, we get:
$$f = A \cdot (x - y) + B \cdot (y - z) - A \cdot (z - x) - B \cdot (z - x).$$
$$f = A \cdot (x - y) - A \cdot (z - x) + B \cdot (y - z) - B \cdot (z - x).$$
$$f = A \cdot (x - y + x - z) + B \cdot (y - z + x - z).$$
$$f = A \cdot (2x - y - z) + B \cdot (y + x - 2z).$$

3. We need to show that there exists a polynomial $B$ such that $A = \frac{B \cdot (y + x - 2z) - f}{y - 2x + z}$ is a polynomial.

4. Consider $f$ as a polynomial in $y$ and perform the division of $f$ by $(y + x - 2z)$. We get:
$$f = g \cdot (y + x - 2z) + h,$$
where $h$ is a polynomial in $x$ and $z$.

5. From the condition $f(w, w, w) = 0$, we have:
$$0 = f(w, w, w) = g(w, w, w) \cdot (w + w - 2w) + h(w, w).$$
This implies that $h(w, w) = 0$ for every real $w$.

6. It is well-known that $x - z$ divides $h$, so we let $h = s \cdot (x - z)$, where $s$ is a polynomial in $x$ and $z$.

7. Therefore, we have:
$$A = -g + \frac{B \cdot (y + x - 2z) - s \cdot (x - z)}{y - 2x + z}.$$

8. Let $B = B_1 \cdot (y - 2x + z) + \frac{s}{3}$. Then we have:
$$A = -g + B_1 \cdot (y + x - 2z) + \frac{\frac{s}{3} \cdot (y + x - 2z) - s \cdot (x - z)}{y - 2x + z}.$$
Simplifying, we get:
$$A = -g + B_1 \cdot (y + x - 2z) + s,$$
which is clearly a polynomial.

9. Since there are infinitely many polynomials $B_1$ we can choose, there is no polynomial $f$ such that $A$, $B$, and $C$ can be uniquely determined.