`(x+y)^(2)-z^(2)=-9, (y+z)^(2)-x^(2)=15, (z+x)^(2)-y^(2)=3`

Options

A. (2,3,4)
B. (1,-1,-3)
C. `-1,1,3`
D. none

Answer

To solve the system of equations given by:

1. \((x+y)^2 - z^2 = -9\)
2. \((y+z)^2 - x^2 = 15\)
3. \((z+x)^2 - y^2 = 3\)

we will follow these steps:

### Step 1: Expand the equations

Start by expanding each equation using the identity \((a+b)^2 = a^2 + b^2 + 2ab\).

1. **First equation**:
\[ (x+y)^2 - z^2 = -9 \implies x^2 + y^2 + 2xy - z^2 = -9 \]
Rearranging gives:
\[ x^2 + y^2 + 2xy - z^2 + 9 = 0 \quad \text{(Equation 1)} \]

2. **Second equation**:
\[ (y+z)^2 - x^2 = 15 \implies y^2 + z^2 + 2yz - x^2 = 15 \]
Rearranging gives:
\[ -x^2 + y^2 + z^2 + 2yz - 15 = 0 \quad \text{(Equation 2)} \]

3. **Third equation**:
\[ (z+x)^2 - y^2 = 3 \implies z^2 + x^2 + 2zx - y^2 = 3 \]
Rearranging gives:
\[ -y^2 + z^2 + x^2 + 2zx - 3 = 0 \quad \text{(Equation 3)} \]

### Step 2: Add Equations

Now, we will add the first and second equations:

\[ (x^2 + y^2 + 2xy - z^2 + 9) + (-x^2 + y^2 + z^2 + 2yz - 15) = 0 \]
This simplifies to:
\[ 2y^2 + 2xy + 2yz - 6 = 0 \]
Dividing by 2 gives:
\[ y^2 + xy + yz - 3 = 0 \quad \text{(Equation 4)} \]

Next, add the first and third equations:

\[ (x^2 + y^2 + 2xy - z^2 + 9) + (-y^2 + z^2 + x^2 + 2zx - 3) = 0 \]
This simplifies to:
\[ 2x^2 + 2xy + 2zx + 6 = 0 \]
Dividing by 2 gives:
\[ x^2 + xy + zx + 3 = 0 \quad \text{(Equation 5)} \]

### Step 3: Solve for variables

Now, we have two new equations (4 and 5). We can express \(y\) in terms of \(x\) and \(z\) using these equations.

From Equation 4:
\[ y^2 + xy + yz = 3 \implies y(y + x + z) = 3 \]

From Equation 5:
\[ x^2 + xy + zx = -3 \implies x^2 + yx + zx = -3 \]

### Step 4: Substitute and Solve

Assuming \(y = -x\) (from the symmetry of the equations), substitute \(y\) into the equations:

1. Substitute \(y = -x\) into Equation 1:
\[ (x - x)^2 - z^2 = -9 \implies -z^2 = -9 \implies z^2 = 9 \implies z = \pm 3 \]

2. Substitute \(z = 3\) back into Equation 1:
\[ (x - x)^2 - 3^2 = -9 \implies -9 = -9 \quad \text{(True)} \]

3. Substitute \(z = -3\) back into Equation 1:
\[ (x - x)^2 - (-3)^2 = -9 \implies -9 = -9 \quad \text{(True)} \]

Thus, we have two cases for \(z\): \(z = 3\) or \(z = -3\). For both cases, we find \(y = -x\).

### Final Values

The solutions for the variables are:
- For \(z = 3\), \(y = -x\)
- For \(z = -3\), \(y = -x\)