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Q. Let $ax ^{3}+ bx ^{2}+ cx + d =\begin{vmatrix}5 x & x+1 & x-1 \\ x-3 & -2 x & x+2 \\ x+3 & x-4 & 5 x\end{vmatrix}$ where $a , b , c$ are constants, then the value of $d$ is

KCETKCET 2022

Solution:

Put $x=0$ we get $d=\begin{vmatrix}0 & 1 & -1 \\ -3 & 0 & 2 \\ 3 & -4 & 0\end{vmatrix}$
$=-1(0-6)-1(12-0)=6-12=-6$