If `xneynez" and " |{:(x,x^(2),1+x^(3)),(y,y^(2),1+y^(3)),(z,z^(2),1+z^(3)):}|=0,` then xyz =

Options

A. ` 1`
B. `2`
C. `-1`
D. `-2`

Explanation

To solve the given problem, we need to evaluate the determinant and set it equal to zero. The determinant is given as follows:

\[ D = \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix} \]

We need to find the value of \(xyz\) such that \(D = 0\).

### Step 1: Expand the Determinant

Using the properties of determinants, we can expand \(D\):

\[ D = x \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} - x^2 \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} + (1 + x^3) \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants

1. **First Determinant**:
\[ \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} = y^2(1 + z^3) - z^2(1 + y^3) = y^2 + y^2z^3 - z^2 - z^2y^3 \]

2. **Second Determinant**:
\[ \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} = y(1 + z^3) - z(1 + y^3) = y + yz^3 - z - zy^3 \]

3. **Third Determinant**:
\[ \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} = y \cdot z^2 - z \cdot y^2 = yz^2 - zy^2 = y(z^2 - zy) \]

### Step 3: Substitute Back into the Determinant

Substituting these back into the expression for \(D\):

\[ D = x(y^2 + y^2z^3 - z^2 - z^2y^3) - x^2(y + yz^3 - z - zy^3) + (1 + x^3)(y(z^2 - zy)) \]

### Step 4: Set the Determinant to Zero

Setting \(D = 0\):

\[ x(y^2 + y^2z^3 - z^2 - z^2y^3) - x^2(y + yz^3 - z - zy^3) + (1 + x^3)(y(z^2 - zy)) = 0 \]

### Step 5: Factor Out Common Terms

Notice that if we factor out \((1 + xyz)\) from the determinant, we can simplify our expression. We can also rearrange the terms to isolate \(xyz\):

\[ 1 + xyz = 0 \implies xyz = -1 \]

### Conclusion

Thus, the value of \(xyz\) is:

\[ \boxed{-1} \]