Let `Delta=|{:(,1,x,x^(2)),(,x^(2),1,x),(,x,x^(2),1):}|`,then

Options

A. `1-x^(2)"is a factor of "Delta`
B. `(1-x)^(2) "is a factor of "Delta`
C. `Delta(x)=0 "has 4 real roots"`
D. `Delta'(1)=0`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} \), we will calculate it step by step.

### Step 1: Write down the determinant
We start with the determinant as given:
\[ \Delta = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} \]

### Step 2: Expand the determinant using the first row
We can use the first row to expand the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & x \\ x^2 & 1 \end{vmatrix} - x \cdot \begin{vmatrix} x^2 & x \\ x & 1 \end{vmatrix} + x^2 \cdot \begin{vmatrix} x^2 & 1 \\ x & x^2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now we will calculate each of the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} 1 & x \\ x^2 & 1 \end{vmatrix} = (1)(1) - (x)(x^2) = 1 - x^3 \]

2. For the second determinant:
\[ \begin{vmatrix} x^2 & x \\ x & 1 \end{vmatrix} = (x^2)(1) - (x)(x) = x^2 - x^2 = 0 \]

3. For the third determinant:
\[ \begin{vmatrix} x^2 & 1 \\ x & x^2 \end{vmatrix} = (x^2)(x^2) - (1)(x) = x^4 - x \]

### Step 4: Substitute back into the determinant
Now substituting back into the expression for \(\Delta\):
\[ \Delta = 1 \cdot (1 - x^3) - x \cdot 0 + x^2 \cdot (x^4 - x) \]
This simplifies to:
\[ \Delta = 1 - x^3 + x^2(x^4 - x) = 1 - x^3 + x^6 - x^3 \]
Combining like terms gives:
\[ \Delta = x^6 - 2x^3 + 1 \]

### Step 5: Final expression
Thus, the final expression for the determinant is:
\[ \Delta = x^6 - 2x^3 + 1 \]