If `|{:(x,x^2,1+x^3),(y,y^2,1+y^3),(z, z^2,1+z^3):}|=0` then relation of `x,y` and `z` is

Options

A. `ane1`
B. `a=1`
C. `a=0`
D. `a=-1`

Answer

To solve the problem, we need to find the relation between \( x, y, \) and \( z \) given that the determinant

\[ D = \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix} = 0. \]

### Step 1: Write the Determinant
We start with the determinant:

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

### Step 2: Expand the Determinant
Using properties of determinants, we can separate the last column:

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

### Step 3: Factor Out Common Terms
We can factor out \( x, y, z \) from the respective rows:

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

### Step 4: Evaluate the Determinants
The first determinant simplifies to:

\[ \begin{vmatrix} 1 & x & 1 \\ 1 & y & 1 \\ 1 & z & 1 \end{vmatrix} = 0 \quad \text{(since rows are linearly dependent)}. \]

The second determinant is a Vandermonde determinant:

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

### Step 5: Set the Determinant to Zero
Thus, we have:

\[ D = xyz \cdot 0 + (y-x)(z-x)(z-y) = 0. \]

### Step 6: Analyze the Factors
The product \( (y-x)(z-x)(z-y) = 0 \) implies that at least one of the factors must be zero:

1. \( y - x = 0 \) → \( y = x \)
2. \( z - x = 0 \) → \( z = x \)
3. \( z - y = 0 \) → \( z = y \)

### Conclusion
Thus, the relations between \( x, y, \) and \( z \) can be summarized as:

- \( x = y \)
- \( y = z \)
- \( z = x \)