Let $a, b, c \in \mathbb{R}$ be such that
$$a + b + c = a^2 + b^2 + c^2 = 1, \hspace{8px} a^3 + b^3 + c^3 \neq 1.$$
We say that a function $f$ is a [i]Palić function[/i] if $f: \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous and satisfies
$$f(x) + f(y) + f(z) = f(ax + by + cz) + f(bx + cy + az) + f(cx + ay + bz)$$
for all $x, y, z \in \mathbb{R}.$
Prove that any Palić function is infinitely many times differentiable and find all Palić functions.
1. **Given Conditions and Definitions:**
   - Let \( a, b, c \in \mathbb{R} \) such that:
     \[
     a + b + c = 1, \quad a^2 + b^2 + c^2 = 1, \quad a^3 + b^3 + c^3 \neq 1
     \]
   - A function \( f \) is a *Palić function* if \( f: \mathbb{R} \rightarrow \mathbb{R} \) is continuous and satisfies:
     \[
     f(x) + f(y) + f(z) = f(ax + by + cz) + f(bx + cy + az) + f(cx + ay + bz)
     \]
     for all \( x, y, z \in \mathbb{R} \).

2. **Initial Setup:**
   - Let \( F' = f \) with \( F(0) = 0 \). Integrate the given functional equation from \( 0 \) to \( 1 \) with respect to \( x \):
     \[
     \int_0^1 f(x) \, dx + f(y) + f(z) = \int_0^1 \left( f(ax + by + cz) + f(bx + cy + az) + f(cx + ay + bz) \right) \, dx
     \]

3. **Integration and Differentiation:**
   - Integrating the left-hand side:
     \[
     \int_0^1 f(x) \, dx + f(y) + f(z) = F(1) + f(y) + f(z)
     \]
   - For the right-hand side, we use the linearity of integration:
     \[
     \int_0^1 f(ax + by + cz) \, dx + \int_0^1 f(bx + cy + az) \, dx + \int_0^1 f(cx + ay + bz) \, dx
     \]
   - By the change of variables and symmetry, we get:
     \[
     \frac{F(ax + by + cz)}{a} + \frac{F(bx + cy + az)}{b} + \frac{F(cx + ay + bz)}{c} \bigg|_0^1
     \]

4. **Differentiability:**
   - Keeping \( z \) fixed and treating \( y \) as a variable, we infer that \( f \) is differentiable.
   - Differentiate the original functional equation with respect to \( x \), \( y \), and \( z \):
     \[
     f'(x) = af'(ax + by + cz) + bf'(bx + cy + az) + cf'(cx + ay + bz)
     \]
     \[
     f'(y) = bf'(ax + by + cz) + cf'(bx + cy + az) + af'(cx + ay + bz)
     \]
     \[
     f'(z) = cf'(ax + by + cz) + af'(bx + cy + az) + bf'(cx + ay + bz)
     \]

5. **Induction and Higher Differentiability:**
   - Adding these equations, we see that \( f' \) satisfies the same condition as \( f \). By induction, \( f \) is infinitely differentiable.

6. **Second Differentiation:**
   - Differentiate the equation with respect to \( y \):
     \[
     0 = abf''(ax + by + cz) + bcf''(bx + cy + az) + caf''(cx + ay + bz)
     \]
   - Differentiate further with respect to \( x \) and set \( x = y = z \):
     \[
     0 = abcf'''(x)
     \]

7. **Conclusion:**
   - Since \( abc \neq 0 \) (otherwise \( a^3 + b^3 + c^3 = 1 \) would be contradicted), we have \( f'''(x) = 0 \).
   - Therefore, \( f \) is a quadratic function. Conversely, any quadratic function satisfies the given conditions.

The final answer is \( \boxed{ f(x) = Ax^2 + Bx + C } \) where \( A, B, C \in \mathbb{R} \).