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

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

we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant as given:

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

### Step 2: Apply row operations
We can simplify the determinant by performing row operations. We will subtract the third row from the first and second rows:

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

This gives us:

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

This simplifies to:

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

### Step 3: Factor out common terms
Notice that \(x^2 - z^2\) and \(y^2 - z^2\) can be factored:

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

Thus, we can factor out \(x - z\) from the first row and \(y - z\) from the second row:

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

### Step 4: Expand the determinant
Now we can expand the determinant along the first column (which has zeros):

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

Calculating this 2x2 determinant:

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

### Step 5: Combine the results
Now substituting back, we have:

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

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

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