The value of the determinant `|{:(1,x,x^2),(1,y,y^2),(1,z,z^2):}|` is equal to

Options

A. `(x-y)(y-z)(z-x)`
B. `(x-y)(y-z)(z-x)(x+y+z)`
C. `(x+y+z)`
D. `(x-y)(y-z)(z-x)(xy+yz+zx)`

Answer

To find the value of the determinant

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

we will use elementary row operations to simplify the determinant.

### Step 1: Perform Row Operations
We will subtract the first row from the second and third rows:

\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1 \]

This gives us:

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

### Step 2: Simplify the Second and Third Rows
Notice that \(y^2 - x^2\) and \(z^2 - x^2\) can be factored using the difference of squares:

\[ y^2 - x^2 = (y - x)(y + x) \quad \text{and} \quad z^2 - x^2 = (z - x)(z + x) \]

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} 1 & x & x^2 \\ 0 & y - x & (y - x)(y + x) \\ 0 & z - x & (z - x)(z + x) \end{vmatrix} \]

### Step 3: Factor Out Common Terms
We can factor out \(y - x\) from the second row and \(z - x\) from the third row:

\[ D = (y - x)(z - x) \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & y + x \\ 0 & 1 & z + x \end{vmatrix} \]

### Step 4: Expand the Remaining Determinant
Now, we can expand the remaining determinant:

\[ D = (y - x)(z - x) \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & y + x \\ 0 & 1 & z + x \end{vmatrix} \]

The determinant simplifies to:

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

### Final Result
Thus, the value of the determinant is:

\[ D = (y - x)(z - x)(y - z) \]