$\left| {\begin{array}{cccccccccccccccccccc}{{x^2} - x + 1}&{x - 1}\\{x + 1}&{x + 1}\end{array}} \right|$
$\left| {\begin{array}{cccccccccccccccccccc}{{x^2} - x + 1}&{x - 1}\\{x + 1}&{x + 1}\end{array}} \right|$
Official Solution
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We have, $\left| {\begin{array}{cccccccccccccccccccc}{{x^2} - x + 1}&{x - 1}\\{x + 1}&{x + 1}\end{array}} \right| = \left| {\begin{array}{cccccccccccccccccccc}{{x^2} - 2x + 2}&{x - 1}\\0&{x + 1}\end{array}} \right|$
$= \left( {{x^2} - 2x + 2} \right) \cdot (x + 1) - (x - 1) \cdot 0$
$= {x^3} - 2{x^2} + 2x + {x^2} - 2x + 2$
$= {x^3} - {x^2} + 2$
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