Question

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

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

Answer 1

(b) 0

$$\begin{gathered} \mathrm{Let}\mathrm{\Delta }=\left|x^2+3xx-1x+3 \\ x+1-2xx-4 \\ x-3x+43x\right| \\ =\left(x^2+3x\right)\left|-2xx-4 \\ x+43x\right|-\left(x-1\right)\left|x+1x-4 \\ x-33x\right|+\left(x+3\right)\left|x+1-2x \\ x-3x+4\right| \\ =\left(x^2+3x\right)\left(-6x-x^2+16\right)-\left(x-1\right)\left(3x^2+3x-x^2+7x-12\right)+\left(x+3\right)\left(x^2+5x+4+2x^2-6x\right) \\ =-7x^4+16x^2+48x+21x^3+8x^2-22x-2x^3-12+8x^2+x+3x^3+12 \\ =-7x^4+22x^3+32x^2+27x+0 \\ \mathrm{But}x\mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}a{x}^4+b{x}^3+c{x}^2+dx+e. \\ \Rightarrow e=0 \end{gathered}$$