If `a^(2) + b^(2) +c^(2)=2 (a-b +c)-3` then find `a-b + c`= ?

Answer

To solve the equation \( a^2 + b^2 + c^2 = 2(a - b + c) - 3 \) and find the value of \( a - b + c \), we will follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ a^2 + b^2 + c^2 = 2(a - b + c) - 3 \]
Distribute the 2 on the right-hand side:
\[ a^2 + b^2 + c^2 = 2a - 2b + 2c - 3 \]

### Step 2: Move all terms to one side
Rearranging the equation gives us:
\[ a^2 + b^2 + c^2 - 2a + 2b - 2c + 3 = 0 \]

### Step 3: Group the terms
Now, we can group the terms to help us identify perfect squares:
\[ a^2 - 2a + b^2 + 2b + c^2 - 2c + 3 = 0 \]

### Step 4: Complete the square for each variable
We will complete the square for each variable:
1. For \( a^2 - 2a \):
\[ a^2 - 2a = (a - 1)^2 - 1 \]
2. For \( b^2 + 2b \):
\[ b^2 + 2b = (b + 1)^2 - 1 \]
3. For \( c^2 - 2c \):
\[ c^2 - 2c = (c - 1)^2 - 1 \]

Substituting these back into the equation gives:
\[ ((a - 1)^2 - 1) + ((b + 1)^2 - 1) + ((c - 1)^2 - 1) + 3 = 0 \]
This simplifies to:
\[ (a - 1)^2 + (b + 1)^2 + (c - 1)^2 = 0 \]

### Step 5: Solve for each variable
Since the sum of squares equals zero, each square must individually equal zero:
\[ (a - 1)^2 = 0 \implies a - 1 = 0 \implies a = 1 \]
\[ (b + 1)^2 = 0 \implies b + 1 = 0 \implies b = -1 \]
\[ (c - 1)^2 = 0 \implies c - 1 = 0 \implies c = 1 \]

### Step 6: Calculate \( a - b + c \)
Now substitute the values of \( a \), \( b \), and \( c \) into \( a - b + c \):
\[ a - b + c = 1 - (-1) + 1 = 1 + 1 + 1 = 3 \]

### Final Answer
Thus, the value of \( a - b + c \) is:
\[ \boxed{3} \]