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

Options

A. all roots real
B. one real, two complex
C. three real roots a,b,c
D. none

Answer

To solve the equation \((x-a)^{3} + (x-b)^{3} + (x-c)^{3} = 0\), we need to analyze the function and its properties. Here’s a step-by-step solution:

### Step 1: Define the function
Let \( f(x) = (x-a)^{3} + (x-b)^{3} + (x-c)^{3} \).

### Step 2: Differentiate the function
To understand the behavior of the function, we differentiate it:
\[ f'(x) = 3(x-a)^{2} + 3(x-b)^{2} + 3(x-c)^{2} \]
This simplifies to:
\[ f'(x) = 3\left((x-a)^{2} + (x-b)^{2} + (x-c)^{2}\right) \]

### Step 3: Analyze the derivative
Each term \((x-a)^{2}\), \((x-b)^{2}\), and \((x-c)^{2}\) is a square term, which means it is always non-negative. Therefore, \(f'(x) \geq 0\) for all \(x\).

### Step 4: Determine the nature of the function
Since \(f'(x) \geq 0\), the function \(f(x)\) is non-decreasing. This means that as \(x\) increases, \(f(x)\) either increases or remains constant but never decreases.

### Step 5: Find the roots
Because \(f(x)\) is a non-decreasing function, it can intersect the x-axis at most once. This means there is at most one real root.

### Step 6: Conclusion about the roots
Since the function is continuous and non-decreasing, it will cross the x-axis at only one point, indicating that there is exactly one real root. The remaining roots, if any, must be complex.

### Final Answer
The equation \((x-a)^{3} + (x-b)^{3} + (x-c)^{3} = 0\) has **one real root and two complex roots**.

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