evaluate: `|(x^(2)-x+1, x-1),(x+1,x+1)|`

Explanation

To evaluate the determinant

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

we will follow these steps:

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

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

### Step 2: Apply the determinant formula
The formula for a 2x2 determinant is given by:

\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]

In our case, \( a = x^2 - x + 1 \), \( b = x - 1 \), \( c = x + 1 \), and \( d = x + 1 \).

### Step 3: Calculate \( ad \) and \( bc \)
Now we will calculate \( ad \) and \( bc \):

1. Calculate \( ad \):
\[ ad = (x^2 - x + 1)(x + 1) = x^3 + x^2 - x^2 - x + x + 1 = x^3 + 1 \]

2. Calculate \( bc \):
\[ bc = (x - 1)(x + 1) = x^2 - 1 \]

### Step 4: Substitute into the determinant formula
Now we substitute \( ad \) and \( bc \) back into the determinant formula:

\[ D = ad - bc = (x^3 + 1) - (x^2 - 1) \]

### Step 5: Simplify the expression
Now we simplify the expression:

\[ D = x^3 + 1 - x^2 + 1 = x^3 - x^2 + 2 \]

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

\[ D = x^3 - x^2 + 2 \]

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