If x,y,z are all distinct and
`|(x,x^2,1+x^3),(y,y^2,1+y^3),(z,z^2,1+z^3)|=0`
then the value of xyz is

Options

A. `-2`
B. `-1`
C. `-3`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant given by:

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

Given that \( |D| = 0 \), we can use properties of determinants to simplify this expression.

### Step 1: Rewrite the determinant
We can rewrite the third column as follows:

\[ 1 + x^3 = 1 + x \cdot x^2 \]

This allows us to express the determinant as:

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

### Step 2: Factor out common terms
We can factor out \( x, y, z \) from the second determinant:

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

The second determinant simplifies to zero since all rows are identical.

### Step 3: Evaluate the first determinant
Now we need to evaluate the first determinant:

\[ D_1 = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} \]

Using the determinant property, we can expand this determinant. The determinant can be calculated as follows:

\[ D_1 = x(y^2 - z^2) - y(x^2 - z^2) + z(x^2 - y^2) \]

### Step 4: Set the determinant to zero
Since we know \( |D| = 0 \), we have:

\[ D_1 = 0 \]

This implies that the rows of the determinant are linearly dependent.

### Step 5: Solve for \( xyz \)
The condition for linear dependence for the determinant \( D_1 \) leads us to conclude that:

\[ \frac{x}{1} = \frac{y}{1} = \frac{z}{1} \]

This means that \( x, y, z \) must satisfy a polynomial equation derived from the determinant being zero.

### Step 6: Find the product \( xyz \)
From the properties of roots of polynomials, we can conclude that the product \( xyz \) must equal -1, based on the structure of the polynomial formed by the roots \( x, y, z \).

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

\[ \boxed{-1} \]