If `x+y+z=0`, then `x^(2)+xy+y^(2)` equals

Options

A. `y^(2)+yz+z^(2)`
B. `y^(2)-yz+z^(2)`
C. `z^(2)-xy`
D. `z^(2)+zx+x^(2)`

Answer

To solve the problem, we need to find the value of \( x^2 + xy + y^2 \) given that \( x + y + z = 0 \).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ x + y + z = 0 \]

2. **Rearrange the equation to express \( x + y \)**:
\[ x + y = -z \]

3. **Substitute \( x + y \) into the expression \( x^2 + xy + y^2 \)**:
We can rewrite \( x^2 + xy + y^2 \) by adding and subtracting \( xy \):
\[ x^2 + xy + y^2 = x^2 + 2xy + y^2 - xy \]

4. **Recognize the algebraic identity**:
The expression \( x^2 + 2xy + y^2 \) can be rewritten using the square of a binomial:
\[ x^2 + 2xy + y^2 = (x + y)^2 \]
Therefore, we can rewrite our expression as:
\[ x^2 + xy + y^2 = (x + y)^2 - xy \]

5. **Substitute \( x + y \) with \(-z\)**:
Now substituting \( x + y = -z \):
\[ (x + y)^2 = (-z)^2 = z^2 \]
So we have:
\[ x^2 + xy + y^2 = z^2 - xy \]

6. **Final expression**:
Thus, the value of \( x^2 + xy + y^2 \) is:
\[ x^2 + xy + y^2 = z^2 - xy \]

### Conclusion:
The expression \( x^2 + xy + y^2 \) simplifies to \( z^2 - xy \).