If `a^(2) + b^(2) + c^(2) = 2(a +2b -2c)-9` then find `a+b+c`=?

Answer

To solve the equation \( a^2 + b^2 + c^2 = 2(a + 2b - 2c) - 9 \), we will follow these steps:

### Step 1: Rearrange the equation
Start by expanding the right side of the equation:

\[ a^2 + b^2 + c^2 = 2a + 4b - 4c - 9 \]

### Step 2: Move all terms to one side
Rearranging gives us:

\[ a^2 + b^2 + c^2 - 2a - 4b + 4c + 9 = 0 \]

### Step 3: Group the terms
Now, we will group the terms in a way that allows us to complete the square:

\[ (a^2 - 2a) + (b^2 - 4b) + (c^2 + 4c) + 9 = 0 \]

### Step 4: Complete the square for each variable
- For \( a^2 - 2a \), we can complete the square:
\[ a^2 - 2a = (a - 1)^2 - 1 \]

- For \( b^2 - 4b \):
\[ b^2 - 4b = (b - 2)^2 - 4 \]

- For \( c^2 + 4c \):
\[ c^2 + 4c = (c + 2)^2 - 4 \]

### Step 5: Substitute back into the equation
Substituting these completed squares back into the equation gives:

\[ ((a - 1)^2 - 1) + ((b - 2)^2 - 4) + ((c + 2)^2 - 4) + 9 = 0 \]

### Step 6: Simplify the equation
Now simplify:

\[ (a - 1)^2 + (b - 2)^2 + (c + 2)^2 - 1 - 4 - 4 + 9 = 0 \]

This simplifies to:

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

### Step 7: Solve for each variable
Since the sum of squares equals zero, each individual square must also equal zero:

1. \( (a - 1)^2 = 0 \) → \( a - 1 = 0 \) → \( a = 1 \)
2. \( (b - 2)^2 = 0 \) → \( b - 2 = 0 \) → \( b = 2 \)
3. \( (c + 2)^2 = 0 \) → \( c + 2 = 0 \) → \( c = -2 \)

### Step 8: Calculate \( a + b + c \)
Now, we find \( a + b + c \):

\[ a + b + c = 1 + 2 - 2 = 1 \]

### Final Answer
Thus, the value of \( a + b + c \) is:

\[ \boxed{1} \]