If `x+y+z=0` then what is the value of
`(1)/(x^(2)+y^(2)-z^(2)) +(1)/(y^(2)+z^(2)-x^(2)) +(1)/(z^(2)+x^(2)-y^(2))`?

Options

A. `(1)/(x^(2)+y^(2)+z^(2))`
B. 1
C. `-1`
D. 0

Answer

To solve the problem, we need to evaluate the expression given the condition \( x + y + z = 0 \). The expression we need to simplify is:

\[ \frac{1}{x^2 + y^2 - z^2} + \frac{1}{y^2 + z^2 - x^2} + \frac{1}{z^2 + x^2 - y^2} \]

### Step 1: Use the condition \( x + y + z = 0 \)

From the condition \( x + y + z = 0 \), we can express \( z \) in terms of \( x \) and \( y \):

\[ z = - (x + y) \]

### Step 2: Substitute \( z \) into the expression

Now, we substitute \( z \) into each term of the expression.

#### First term:

\[ x^2 + y^2 - z^2 = x^2 + y^2 - (- (x + y))^2 = x^2 + y^2 - (x^2 + 2xy + y^2) = x^2 + y^2 - x^2 - 2xy - y^2 = -2xy \]

Thus, the first term becomes:

\[ \frac{1}{x^2 + y^2 - z^2} = \frac{1}{-2xy} \]

#### Second term:

\[ y^2 + z^2 - x^2 = y^2 + (- (x + y))^2 - x^2 = y^2 + (x^2 + 2xy + y^2) - x^2 = 2y^2 + 2xy \]

Thus, the second term becomes:

\[ \frac{1}{y^2 + z^2 - x^2} = \frac{1}{2y^2 + 2xy} = \frac{1}{2y(y + x)} \]

#### Third term:

\[ z^2 + x^2 - y^2 = (- (x + y))^2 + x^2 - y^2 = (x^2 + 2xy + y^2) + x^2 - y^2 = 2x^2 + 2xy \]

Thus, the third term becomes:

\[ \frac{1}{z^2 + x^2 - y^2} = \frac{1}{2x^2 + 2xy} = \frac{1}{2x(x + y)} \]

### Step 3: Combine all terms

Now we combine all three terms:

\[ \frac{1}{-2xy} + \frac{1}{2y(y + x)} + \frac{1}{2x(x + y)} \]

### Step 4: Find a common denominator

The common denominator for these fractions is \( -2xy \cdot 2y(y + x) \cdot 2x(x + y) \).

### Step 5: Simplify the expression

After finding the common denominator, we can combine the numerators. However, we notice that due to the symmetry and the condition \( x + y + z = 0 \), the entire expression simplifies to zero.

### Final Answer

Thus, the value of the expression is:

\[ \boxed{0} \]