Aspire Faculty ID #4277 · Topic: NIMCET 2017 · Just now
NIMCET 2017

The equation $(x-a)^3+(x-b)^3+(x-c)^3=0$ has

Solution


Let f(x) = (x – a)3 + (x – b)3 + (x – c)3.
Then f'(x) = 3{(x – a)2 + (x – b)2 + (x –c)2}
clearly , f'(x) > 0 for all x.
so, f'(x) = 0 has no real roots.
Hence, f(x) = 0 has two imaginary and one real root
Let the given equation be: 
$f(x) = (x - a)^3 + (x - b)^3 + (x - c)^3 = 0$ 

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**.
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