Δ x = ∣ ∣ ∣ ∣ ∣ ∣ x 2 x + 1 x 2 + x x 2 2 x 2 − 1 x 3 + x 2 2 x + 1 4 x + 2 2 x 2 + 3 x + 2 ∣ ∣ ∣ ∣ ∣ ∣ Then find value of : x = 1 ∑ 1 0 Δ x
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Perform the following row operations: Replace R3 by R3-R1. After that replace R2 by R2-2R1. After that replace R3 by R3- x R1. After that expand the determinant along the third row to get the value of determinant as − x - x 2 . After that we get the value of summation as − 4 4 0 .
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D=-x-x^2 So D = -440