If a,b,c are real, then `a(a - b) + b ( b - c) + c ( c - a) = 0 `, only if

Options

A. `a + b + c = 0 `
B. ` a = b = c `
C. ` a = b " or " b = c " or " c = a `
D. ` a - b -c = 0 `

Answer

To solve the equation \( a(a - b) + b(b - c) + c(c - a) = 0 \), we will simplify and analyze it step by step.

### Step 1: Expand the Expression
We start by expanding the left-hand side of the equation:

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

Expanding each term:

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

So, we rewrite the equation as:

\[ a^2 + b^2 + c^2 - ab - bc - ac = 0 \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:

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

### Step 3: Analyzing the Expression
The equation \( a^2 + b^2 + c^2 = ab + bc + ac \) can be interpreted in terms of the squares of differences. We can rewrite it as:

\[ \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0 \]

### Step 4: Setting Each Square to Zero
Since the sum of squares is zero, each individual square must also be zero:

\[ (a-b)^2 = 0, \quad (b-c)^2 = 0, \quad (c-a)^2 = 0 \]

This implies:

\[ a - b = 0, \quad b - c = 0, \quad c - a = 0 \]

### Step 5: Conclusion
From the above equations, we conclude that:

\[ a = b = c \]

Thus, the original equation \( a(a - b) + b(b - c) + c(c - a) = 0 \) holds true if and only if \( a = b = c \).

### Final Answer
The correct condition is: **Option B: \( a = b = c \)**.

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