$f(x) = (x - a)^3 + (x - b)^3 + (x - c)^3 = 0$ 

Differentiate $f(x)$ with respect to $x$ 

$f'(x) = 3(x - a)^2 + 3(x - b)^2 + 3(x - c)^2$ 

$f'(x) = 3\left[(x - a)^2 + (x - b)^2 + (x - c)^2\right]$ 

Step 2: Analyze the Nature of $f'(x)$ 

Since the square of any real number is non-negative: 

$(x - a)^2 \ge 0, \quad (x - b)^2 \ge 0, \quad (x - c)^2 \ge 0 \quad \text{for all } x \in \mathbb{R}$ 

If $a, b, c$ are not all equal, the sum of these three non-negative squares cannot simultaneously be zero for any real $x$: 

 $f'(x) > 0 \quad \text{for all } x \in \mathbb{R}$ 

If $a = b = c$, $f'(x) \ge 0$ (vanishing only at the single point $x = a$). 

In both cases, $f(x)$ is a **strictly increasing function** on $\mathbb{R}$.

Conclusion: 

1. As a cubic polynomial with real coefficients, $f(x) \to -\infty$ as $x \to -\infty$, and $f(x) \to +\infty$ as $x \to +\infty$. 

2. Because $f(x)$ is strictly increasing, it can cross the x-axis **at most once**. 

 Therefore, the equation has **exactly one real root** and **two imaginary (complex conjugate) roots**.