-If `a,b,c in R` then prove that the roots of the equation `1/(x-a)+1/(x-b)+1/(x-c)=0` are always real cannot have roots if `a=b=c`

Options

A. One roots lies in (a,b)
B. One root lies in (b,c)
C. One root must be non-real
D. Three roots are real.

Explanation

To prove that the roots of the equation

\[ \frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 \]

are always real and cannot have roots if \( a = b = c \), we can follow these steps:

### Step 1: Rewrite the Equation
We start by rewriting the equation:

\[ \frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 \]

This can be combined into a single fraction:

\[ \frac{(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)}{(x-a)(x-b)(x-c)} = 0 \]

The numerator must be equal to zero for the entire fraction to be zero, so we focus on:

\[ (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0 \]

### Step 2: Expand the Numerator
Now, we expand the numerator:

\[ (x-b)(x-c) = x^2 - (b+c)x + bc \]
\[ (x-a)(x-c) = x^2 - (a+c)x + ac \]
\[ (x-a)(x-b) = x^2 - (a+b)x + ab \]

Adding these together:

\[ 3x^2 - (2a + 2b + 2c)x + (ab + ac + bc) = 0 \]

### Step 3: Form a Quadratic Equation
This gives us a quadratic equation in the form:

\[ 3x^2 - 2(a+b+c)x + (ab + ac + bc) = 0 \]

### Step 4: Calculate the Discriminant
For the roots to be real, the discriminant of this quadratic must be non-negative. The discriminant \( D \) is given by:

\[ D = B^2 - 4AC \]

Where \( A = 3 \), \( B = -2(a+b+c) \), and \( C = ab + ac + bc \). Thus,

\[ D = [-2(a+b+c)]^2 - 4 \cdot 3 \cdot (ab + ac + bc) \]
\[ D = 4(a+b+c)^2 - 12(ab + ac + bc) \]

### Step 5: Analyze the Discriminant
Now, we need to analyze this discriminant.

1. If \( a, b, c \) are not all equal, then \( D \) will be positive, which means the roots are real and distinct.
2. If \( a = b = c \), then substituting \( a = b = c \) into the discriminant gives:

\[ D = 4(3a)^2 - 12(3a^2) = 36a^2 - 36a^2 = 0 \]

This means the discriminant is zero, indicating that the roots are real but equal.

### Conclusion
Thus, we conclude that:

- The roots of the equation are always real when \( a, b, c \) are not all equal.
- If \( a = b = c \), the equation does not have distinct roots, indicating that the roots cannot be considered as separate.