If `(x+y-z)^(2) +(y+z-x)^(2) +(z+x-y)^(2)=0`
then what is the value of `x+y+z`?

Options

A. `sqrt(3)`
B. `3sqrt(3)`
C. 3
D. 0

Answer

To solve the equation \((x+y-z)^{2} + (y+z-x)^{2} + (z+x-y)^{2} = 0\), we can follow these steps:

### Step 1: Understand the equation
The equation consists of three squared terms added together, and the sum is equal to zero.

### Step 2: Analyze the implications of the equation
Since the sum of squares is zero, each individual square must also be zero. This is because squares of real numbers are non-negative, and the only way their sum can equal zero is if each term is zero.

### Step 3: Set each squared term to zero
1. \( (x+y-z)^{2} = 0 \)
2. \( (y+z-x)^{2} = 0 \)
3. \( (z+x-y)^{2} = 0 \)

From this, we can derive three equations:
1. \( x + y - z = 0 \) (Equation 1)
2. \( y + z - x = 0 \) (Equation 2)
3. \( z + x - y = 0 \) (Equation 3)

### Step 4: Solve the equations
From Equation 1:
\[ x + y = z \]

From Equation 2:
\[ y + z = x \implies z = x - y \]

From Equation 3:
\[ z + x = y \implies z = y - x \]

### Step 5: Equate the expressions for \( z \)
From \( z = x - y \) and \( z = y - x \):
\[ x - y = y - x \]

### Step 6: Simplify the equation
Rearranging gives:
\[ x + x = y + y \implies 2x = 2y \implies x = y \]

### Step 7: Substitute back to find \( z \)
Using \( x = y \) in Equation 1:
\[ x + x = z \implies z = 2x \]

### Step 8: Find \( x + y + z \)
Now, substituting \( y = x \) and \( z = 2x \):
\[ x + y + z = x + x + 2x = 4x \]

### Step 9: Determine the value of \( x + y + z \)
Since all equations must hold true, and the only way for the squares to equal zero is if \( x = y = z = 0 \):
\[ x + y + z = 4(0) = 0 \]

### Final Answer
Thus, the value of \( x + y + z \) is \( \boxed{0} \).

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