If the roots of the equation `(x-b) (x-c) + (x-c) (x-a) + (x-a) (x-b)=0` are equal, then

Options

A. a+b+c =0
B. `a+b omega +c omega^(2)=0`
C. `a-b+c=0`
D. `a+b omega^(2)+c omega =0`

Answer

To solve the equation \((x-b)(x-c) + (x-c)(x-a) + (x-a)(x-b) = 0\) and find the condition for equal roots, we can follow these steps:

### Step 1: Expand the Equation
We start by expanding each term in the equation:

1. \((x-b)(x-c) = x^2 - (b+c)x + bc\)
2. \((x-c)(x-a) = x^2 - (c+a)x + ac\)
3. \((x-a)(x-b) = x^2 - (a+b)x + ab\)

Now, we can combine these expansions:

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

### Step 2: Combine Like Terms
Combine all the \(x^2\) terms, the \(x\) terms, and the constant terms:

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

The coefficient of \(x\) simplifies to:

\[ -(2a + 2b + 2c) = -2(a + b + c) \]

Thus, the equation becomes:

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

### Step 3: Identify Coefficients
From the standard form \(Ax^2 + Bx + C = 0\), we identify:
- \(A = 3\)
- \(B = -2(a+b+c)\)
- \(C = ab + ac + bc\)

### Step 4: Use the Condition for Equal Roots
For the roots to be equal, the discriminant must be zero:

\[ D = B^2 - 4AC = 0 \]

Substituting the values of \(A\), \(B\), and \(C\):

\[ (-2(a+b+c))^2 - 4(3)(ab + ac + bc) = 0 \]

### Step 5: Simplify the Discriminant
Expanding the discriminant:

\[ 4(a+b+c)^2 - 12(ab + ac + bc) = 0 \]

### Step 6: Rearranging the Equation
Rearranging gives us:

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

Dividing through by 4:

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

### Conclusion
The condition for the roots of the given equation to be equal is:

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