If `(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0`, then the values of a, b, and c are ___________.

Answer

To solve the equation \((a-b)^{2} + (b-c)^{2} + (c-a)^{2} = 0\), we can follow these steps:

### Step 1: Understand the properties of squares
Since the expression consists of squares, we know that each square term is non-negative. This means that:
\[ (a-b)^{2} \geq 0, \quad (b-c)^{2} \geq 0, \quad (c-a)^{2} \geq 0 \]

### Step 2: Analyze the sum of squares
The sum of these squares equals zero:
\[ (a-b)^{2} + (b-c)^{2} + (c-a)^{2} = 0 \]
For this sum to be zero, each individual square must also be zero. Therefore, we can set each term to zero:
\[ (a-b)^{2} = 0, \quad (b-c)^{2} = 0, \quad (c-a)^{2} = 0 \]

### Step 3: Solve each equation
From \((a-b)^{2} = 0\), we get:
\[ a - b = 0 \implies a = b \]

From \((b-c)^{2} = 0\), we get:
\[ b - c = 0 \implies b = c \]

From \((c-a)^{2} = 0\), we get:
\[ c - a = 0 \implies c = a \]

### Step 4: Conclude the values of \(a\), \(b\), and \(c\)
From the above equations, we can conclude:
\[ a = b = c \]

Thus, the values of \(a\), \(b\), and \(c\) are equal.

### Final Answer
The values of \(a\), \(b\), and \(c\) are ___________ (they are all equal).

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