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 determinant equation \( |(x, x^2, 1+x^3), (y, y^2, 1+y^3), (z, z^2, 1+z^3)| = 0 \), we can follow these steps:

### 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 the property of determinants, we can expand it:
\[ D = x \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} - x^2 \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} + (1 + x^3) \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:
1. \( \begin{vmatrix} y^2 & 1 + y^3 \\ z^2 & 1 + z^3 \end{vmatrix} = y^2(1 + z^3) - z^2(1 + y^3) = y^2 + y^2z^3 - z^2 - z^2y^3 \)

2. \( \begin{vmatrix} y & 1 + y^3 \\ z & 1 + z^3 \end{vmatrix} = y(1 + z^3) - z(1 + y^3) = y + yz^3 - z - zy^3 \)

3. \( \begin{vmatrix} y & y^2 \\ z & z^2 \end{vmatrix} = y z^2 - z y^2 = y(z^2 - y z) \)

### Step 4: Substitute Back into the Determinant
Substituting these back into \( D \):
\[ D = x(y^2 + y^2z^3 - z^2 - z^2y^3) - x^2(y + yz^3 - z - zy^3) + (1 + x^3)(y(z^2 - yz)) \]

### Step 5: Set the Determinant to Zero
Setting \( D = 0 \) gives us the condition for \( x, y, z \) to be linearly dependent.

### Step 6: Analyze the Result
The determinant being zero implies that the rows (or columns) of the matrix are linearly dependent. This means that the points represented by \( (x, x^2, 1+x^3) \), \( (y, y^2, 1+y^3) \), and \( (z, z^2, 1+z^3) \) are collinear.

### Conclusion
Thus, the relation among \( x, y, z \) is that they must satisfy a certain polynomial relationship that makes these points collinear.