If a + b+ c =2s , thent the value of `(s-a)^(2) + (s-b)^(2) +( s-c)^2` will be

Options

A. `s^2 + a ^2 + b^2 + c^2`
B. `a^2 + b^2 + c^2-s^2`
C. `s^2 -a^2-b^2-c^2`
D. `4s^2 - a^2-b^2-c^2`

Answer

To solve the problem, we need to find the value of \((s-a)^2 + (s-b)^2 + (s-c)^2\) given that \(a + b + c = 2s\).

### Step-by-Step Solution:

1. **Start with the expression**:
\[ (s-a)^2 + (s-b)^2 + (s-c)^2 \]

2. **Expand each square using the identity \((x-y)^2 = x^2 - 2xy + y^2\)**:
\[ (s-a)^2 = s^2 - 2sa + a^2 \]
\[ (s-b)^2 = s^2 - 2sb + b^2 \]
\[ (s-c)^2 = s^2 - 2sc + c^2 \]

3. **Combine the expanded expressions**:
\[ (s-a)^2 + (s-b)^2 + (s-c)^2 = (s^2 - 2sa + a^2) + (s^2 - 2sb + b^2) + (s^2 - 2sc + c^2) \]

4. **Group like terms**:
\[ = 3s^2 - 2s(a + b + c) + (a^2 + b^2 + c^2) \]

5. **Substitute \(a + b + c = 2s\)** into the equation:
\[ = 3s^2 - 2s(2s) + (a^2 + b^2 + c^2) \]
\[ = 3s^2 - 4s^2 + (a^2 + b^2 + c^2) \]

6. **Simplify the expression**:
\[ = -s^2 + (a^2 + b^2 + c^2) \]
\[ = a^2 + b^2 + c^2 - s^2 \]

### Final Result:
Thus, the value of \((s-a)^2 + (s-b)^2 + (s-c)^2\) is:
\[ \boxed{a^2 + b^2 + c^2 - s^2} \]