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# If $ax^3+bx^2+cx+d=\begin{vmatrix}x^2&(x-1)^2&(x-2)\\(x-)^2&(x-2)^2&(x-3)^2\\(x-2)^2&(x-3)^2&(x-4)^2\end{vmatrix}$ then

$\begin{array}{1 1}(A)\;\text{a=1,b=2,c=3,d=-8}\\(B)\;\text{a=-1,b=2,c=3,d=-8}\\(C)\;\text{a=0,b=0,c=0,d=8}\\(D)\;\text{a=0,b=0,c=0,d=-8}\end{array}$

Apply $C_1\rightarrow C_1-C_2$
$C_2\rightarrow C_2-C_3$
$\begin{vmatrix} (2x-1)&(2x-3)&(x-2)^2\\(2x-3)&(2x-5)&(x-3)^2\\(2x-5)&(2x-7)&(x-4)^2\end{vmatrix}$
Apply $R\rightarrow R_1-R_2,R_2\rightarrow R_2-R_3$
$\begin{vmatrix}2&2&2x-5\\2&2&2x-7\\(2x-5)&(2x-7)&(x-4)^2\end{vmatrix}$
Applying $R_1\rightarrow R_1-R_2$
$\begin{vmatrix}0&0&2\\2&2&(2x-7)\\(2x-5)&(2x-7)&(x-4)^2\end{vmatrix}=-8$
$\therefore$ Value of determinant is independent of $x$
$\therefore a=b=c=0$ and $d=-8$
Hence (D) is the correct answer.