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. 0
B. -1
C. -2
D. -3

Answer

To solve the given determinant problem, we start with the determinant:

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

We know that if the determinant \(D = 0\), then the rows of the determinant are linearly dependent.

### Step 1: Expand the determinant

We can expand the determinant using the properties of determinants. The third row can be rewritten as:

\[ 1 + x^3 = 1 + x^3, \quad 1 + y^3 = 1 + y^3, \quad 1 + z^3 = 1 + z^3 \]

Thus, we can rewrite the determinant as:

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

### Step 2: Use properties of determinants

We can perform row operations to simplify the determinant. Subtract the first row from the third row:

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

### Step 3: Factor the third row

Notice that \(x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1)\). Similarly for \(y\) and \(z\):

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

### Step 4: Factor out common terms

Now we can factor out \(x\), \(y\), and \(z\) from the third row:

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

### Step 5: Set the determinant to zero

Since \(D = 0\), we have:

\[ xyz \cdot \text{(some determinant)} = 0 \]

Since \(x\), \(y\), and \(z\) are not equal, the only solution is:

\[ xyz = 0 \]

### Conclusion

Thus, we conclude that:

\[ xyz = 0 \]