If `a^2+b^2+c^2-a b-b c-c a=0,` then find the relation between a,b and c

Options

A. `a=b= - c`
B. `a=b=c`
C. `a= - b=c`
D. `-a=b=c`

Answer

To solve the equation \( a^2 + b^2 + c^2 - ab - bc - ca = 0 \) and find the relationship between \( a \), \( b \), and \( c \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ a^2 + b^2 + c^2 - ab - bc - ca = 0 \]

### Step 2: Multiply the entire equation by 2
To simplify the expression, we can multiply the entire equation by 2:
\[ 2(a^2 + b^2 + c^2 - ab - bc - ca) = 0 \]
This simplifies to:
\[ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 \]

### Step 3: Rearrange the equation
We can rearrange the equation as follows:
\[ 2a^2 - 2ab + 2b^2 - 2bc + 2c^2 - 2ca = 0 \]

### Step 4: Group terms
We can group the terms to form perfect squares:
\[ (a^2 - ab + b^2) + (b^2 - bc + c^2) + (c^2 - ca + a^2) = 0 \]

### Step 5: Recognize the perfect square forms
Notice that:
\[ a^2 - ab + b^2 = \frac{1}{2}((a-b)^2 + a^2 + b^2) \]
Similarly, we can express the other groups:
\[ b^2 - bc + c^2 = \frac{1}{2}((b-c)^2 + b^2 + c^2) \]
\[ c^2 - ca + a^2 = \frac{1}{2}((c-a)^2 + c^2 + a^2) \]

### Step 6: Set each perfect square to zero
Since the sum of squares is zero, each individual square must also be zero:
\[ (a - b)^2 = 0 \]
\[ (b - c)^2 = 0 \]
\[ (c - a)^2 = 0 \]

### Step 7: Solve the equations
From these equations, we can deduce:
1. \( a - b = 0 \) implies \( a = b \)
2. \( b - c = 0 \) implies \( b = c \)
3. \( c - a = 0 \) implies \( c = a \)

### Conclusion
Thus, we conclude that:
\[ a = b = c \]