The value of `|{:(x,x^2-yz,1),(y,y^2-zx,1),(z,z^2-xy,1):}|` is

Options

A. 1
B. -1
C. 0
D. #REF!

Answer

To find the value of the determinant

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

we can follow these steps:

### Step 1: Rewrite the Determinant
We can express the determinant as:

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

### Step 2: Perform Column Operations
We can perform column operations to simplify the determinant. Let's subtract the third column from the second column:

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

This gives us:

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

### Step 3: Factor Out Common Terms
Now, we can factor out the common terms from the second column:

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

### Step 4: Check for Identical Rows
Now, we can see if we can manipulate the determinant further. Notice that if we perform the operation of switching columns, we can see that the determinant can become identical to itself:

If we switch the second and third columns, we have:

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

### Step 5: Identify Identical Determinants
Now, if we switch the first column with the second column, we get:

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

### Step 6: Conclusion
Since we have two identical determinants, the value of the determinant must be zero:

\[ D = 0 \]

Thus, the value of the determinant is:

\[ \boxed{0} \]