If D(x) = begin{vmatrix} x - 1 & (x - 1)^2 & | vedclass

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(D) To find the coefficient of $x$ in the polynomial $D(x)$,we can use the property that the coefficient of $x$ is given by $D'(0)$.First,evaluate $D(

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$0$

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(D) To find the coefficient of $x$ in the polynomial $D(x)$,we can use the property that the coefficient of $x$ is given by $D'(0)$.First,evaluate $D(0)$:$D(0) = \begin{vmatrix} -1 & 1 & 0 \ -1 & 0 & 1 \ 0 & 1 & 1 \end{vmatrix} = -1(0 - 1) - 1(-1 - 0) + 0 = 1 + 1 = 2$.Since $D(x)$ is a polynomial,we can write $D(x) = a_n x^n + \dots + a_1 x + a_0$,where $a_1$ is the coefficient of $x$.By differentiating $D(x)$ with respect to $x$ and evaluating at $x=0$,we find the coefficient.Alternatively,expanding the determinant:$D(x) = (x-1)[x^2(x+1)^3 - (x+1)^2(x+1)^3] - (x-1)^2[(x-1)(x+1)^3 - x(x+1)^3] + x^3[(x-1)(x+1)^2 - x^2(x-1)]$.By simplifying the expression or evaluating the derivative at $0$,we find that the constant term is $2$ and the linear term coefficient is $0$.

If $D(x) = \begin{vmatrix} x - 1 & (x - 1)^2 & x^3 \\ x - 1 & x^2 & (x + 1)^3 \\ x & (x + 1)^2 & (x + 1)^3 \end{vmatrix}$,then the coefficient of $x$ in $D(x)$ is

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