Solve this


Let $\left|\begin{array}{ccc}x^{2}+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x\end{array}\right|=a x^{4}+b x^{3}+c x^{2}+d x+e$

be an identity in $x$, where $a, b, c, d$, e are independent of $x$. Then the value of $e$ is

(a) 4

(b) 0

(c) 1

(d) none of these


(b) 0

Let $\Delta=\mid x^{2}+3 x \quad x-1 \quad x+3$

$x+1 \quad-2 x \quad x-4$

$x-3 \quad x+4 \quad 3 x$

$=\left(x^{2}+3 x\right) \mid-2 x \quad x-4$

$x+4 \quad 3 x|-(x-1)| x+1 \quad x-4$

$x-3 \quad 3 x|+(x+3)| x+1 \quad-2 x$

$x-3 \quad x+4$

$=\left(x^{2}+3 x\right)\left(-6 x-x^{2}+16\right)-(x-1)\left(3 x^{2}+3 x-x^{2}+7 x-12\right)+(x+3)\left(x^{2}+5 x+4+2 x^{2}-6 x\right)$

$=-7 x^{4}+16 x^{2}+48 x+21 x^{3}+8 x^{2}-22 x-2 x^{3}-12+8 x^{2}+x+3 x^{3}+12$

$=-7 x^{4}+22 x^{3}+32 x^{2}+27 x+0$

But $x$ is a root of $a x^{4}+b x^{3}+c x^{2}+d x+e$.

$\Rightarrow e=0$

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