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The value of ∣x2−x+1x−1x+1x+1∣\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} is equal to:

Solution

✅ Correct Option: 3

The value of a 2×22 \times 2 determinant ∣abcd∣\begin{vmatrix} a & b \\ c & d \end{vmatrix} is given by ad−bcad - bc.

Δ=∣x2−x+1x−1x+1x+1∣\Delta = \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}

Multiply the diagonal elements:

Δ=(x2−x+1)(x+1)−(x−1)(x+1)\Delta = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)

We use two standard algebraic identities:

  1. Sum of cubes: (x+1)(x2−x+1)=x3+13(x+1)(x^2 - x + 1) = x^3 + 1^3
  2. Difference of squares: (x−1)(x+1)=x2−12(x-1)(x+1) = x^2 - 1^2

Substitute these back into the expression:

Δ=(x3+1)−(x2−1)\Delta = (x^3 + 1) - (x^2 - 1)

Δ=x3+1−x2+1\Delta = x^3 + 1 - x^2 + 1

Δ=x3−x2+2\Delta = x^3 - x^2 + 2

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