If a x^{4}+b x^{3}+c x^{2}+d x+e = left|begin | vedclass

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(B) Given the equation: $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \ x+1 & -2 x & x-4 \ x-3 & x+4 & 3 x\end{arra

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(B) Given the equation: $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \ x+1 & -2 x & x-4 \ x-3 & x+4 & 3 x\end{array}ight|$.To find the value of $e$,we set $x = 0$ on both sides of the equation.Substituting $x = 0$ into the determinant,we get:$e = \left|\begin{array}{ccc}0+3(0) & 0-1 & 0+3 \ 0+1 & -2(0) & 0-4 \ 0-3 & 0+4 & 3(0)\end{array}ight| = \left|\begin{array}{ccc}0 & -1 & 3 \ 1 & 0 & -4 \ -3 & 4 & 0\end{array}ight|$.Expanding the determinant along the first row:$e = 0(0 - (-16)) - (-1)(0 - 12) + 3(4 - 0)$$e = 0(16) + 1(-12) + 3(4)$$e = 0 - 12 + 12 = 0$.Thus,$e = 0$.

If $a x^{4}+b x^{3}+c x^{2}+d x+e = \left|\begin{array}{ccc}x^{3}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|$,then $e$ is equal to

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