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

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(B) To find the coefficient of $x$ in $\Delta(x)$, we use the property that the coefficient of $x$ is given by $\Delta'(0)$.Given $\Delta(x) = \begin{

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$-2$

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(B) To find the coefficient of $x$ in $\Delta(x)$, we use the property that the coefficient of $x$ is given by $\Delta'(0)$.Given $\Delta(x) = \begin{vmatrix} x-2 & (x-1)^2 & x^3 \ x-1 & x^2 & (x+1)^3 \ x & (x+1)^2 & (x+2)^3 \end{vmatrix}$.Using the property of differentiation of a determinant, $\Delta'(x) = \Delta_1(x) + \Delta_2(x) + \Delta_3(x)$, where $\Delta_i$ is the determinant with the $i$-th row differentiated.Evaluating $\Delta'(0)$ involves calculating the derivative of each row at $x=0$.After performing the differentiation and substituting $x=0$, we obtain $\Delta'(0) = -2$.Thus, the coefficient of $x$ is $-2$.

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

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