A) \[{{x}^{3}}\]
B) \[{{x}^{2}}\]
C) \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]
D) None of these
Correct Answer: B
Solution :
\[\Delta =\frac{1}{abc}\,\left| \,\begin{matrix} {{a}^{3}}+ax & {{a}^{2}}b & {{a}^{2}}c \\ a{{b}^{2}} & {{b}^{3}}+bx & {{b}^{2}}c \\ {{c}^{2}}a & {{c}^{2}}b & {{c}^{3}}+cx \\ \end{matrix}\, \right|\] = \[(abc)\,(ac+ab+bc)=a+b+c\] = \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+x)\times \left| \,\begin{matrix} 1 & 1 & 1 \\ {{b}^{2}} & {{b}^{2}}+x & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{c}^{2}}+x \\ \end{matrix}\, \right|\] {Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\}\] \[=({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+x)\,\left| \,\begin{matrix} 1 & 0 & 0 \\ {{b}^{2}} & x & 0 \\ {{c}^{2}} & 0 & x \\ \end{matrix}\, \right|\]\[{{A}^{2}}=A\,.\,A=\left[ \begin{matrix} \text{2} & -1 \\ -1 & 2 \\ \end{matrix} \right]\,\left[ \begin{matrix} \text{2} & -1 \\ -1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} \text{5} & -4 \\ -4 & 5 \\ \end{matrix} \right]\]\[\left. \begin{align} & {{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \\ & {{C}_{3}}\to {{C}_{3}}-{{C}_{1}} \\ \end{align} \right\}\] = \[{{x}^{2}}({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+x)\]. Hence \[\Delta \] is divisible by \[{{x}^{2}}\] as well as by x.
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