If $ P(x) $ is a polynomial of degree 5 and $$ \alpha = \sum_{i=0}^{6} P(x_i) \left( \prod_{\substack{j=0 j\neq i}}^{6} (x_i - x_j)^{-1} \right), $$ where $ x_0, x_1, \ldots, x_6 $ are distinct points in the interval $[2,3]$, then the value of $ \alpha^2 - \alpha + 1 $ is __________.

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The given expression for $ \alpha $ is a form of the Lagrange interpolation formula. This formula is used to express a polynomial passing through a set of points. Given that $ P(x) $ is a polynomial of degree 5, the expression for $ \alpha $ will essentially sum the values of the polynomial at each of the distinct points $ x_0, x_1, \ldots, x_6 $, weighted by their corresponding Lagrange basis polynomials. The key observation is that this interpolation formula sums the values of the polynomial $ P(x) $ evaluated at distinct points, and since the polynomial $ P(x) $ has degree 5, we know that $ P(x) $ is uniquely determined by these points. The structure of the formula for $ \alpha $ suggests that $ \alpha = 1 $, based on the properties of the Lagrange interpolation and the fact that the sum of Lagrange basis polynomials for a complete set of distinct points equals 1. Thus, $ \alpha = 1 $. Now, we compute $ \alpha^2 - \alpha + 1 $: $$ \alpha^2 - \alpha + 1 = 1^2 - 1 + 1 = 1. $$ Thus, the value of $ \alpha^2 - \alpha + 1 $ is $\boxed{1}$.

If \( P(x) \) is a polynomial of degree 5 and \[ \alpha = \sum_{i=0}^{6} P(x_i) \left( \prod_{\substack{j=0
j\neq i}}^{6} (x_i - x_j)^{-1} \right), \] where \( x_0, x_1, \ldots, x_6 \) are distinct points in the interval \([2,3]\), then the value of \( \alpha^2 - \alpha + 1 \) is __________.

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Correct Answer: 1

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If \( P(x) \) is a polynomial of degree 5 and \[ \alpha = \sum_{i=0}^{6} P(x_i) \left( \prod_{\substack{j=0 j\neq i}}^{6} (x_i - x_j)^{-1} \right), \] where \( x_0, x_1, \ldots, x_6 \) are distinct points in the interval \([2,3]\), then the value of \( \alpha^2 - \alpha + 1 \) is __________.