If `x+ y+ z= 2s` find `(s-x)^(3) + (s-y)^(3) + 3(s-x) (s-y) z=` ?

Options

A. `z^(3)`
B. `-z^(3)`
C. 0
D. `2z^(3)`

Answer

To solve the problem, we need to find the expression \( (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)z \) given that \( x + y + z = 2s \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ x + y + z = 2s \]

2. **Rearrange the equation**:
We can express \( z \) in terms of \( s \), \( x \), and \( y \):
\[ z = 2s - x - y \]

3. **Substitute \( z \) into the expression**:
We need to evaluate:
\[ (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)z \]
Substitute \( z \):
\[ = (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)(2s - x - y) \]

4. **Use the identity for the sum of cubes**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Here, let \( a = s-x \), \( b = s-y \), and \( c = -(2s - x - y) \). Notice that:
\[ a + b + c = (s-x) + (s-y) - (2s - x - y) = 0 \]

5. **Apply the identity**:
Since \( a + b + c = 0 \), we have:
\[ (s-x)^3 + (s-y)^3 + (-(2s - x - y))^3 = 3(s-x)(s-y)(2s - x - y) \]
This simplifies to:
\[ (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)(-(2s - x - y)) = 0 \]

6. **Rearranging the equation**:
Thus, we can rearrange to find:
\[ (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)z = z^3 \]

7. **Final result**:
Therefore, we conclude that:
\[ (s-x)^3 + (s-y)^3 + 3(s-x)(s-y)z = z^3 \]

### Conclusion:
The final answer is:
\[ \boxed{z^3} \]