A) 0
B) \[(a-b)(b-c)(c-a)\]
C) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
D) None of these
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} a-1 & a & bc \\ b-1 & b & ca \\ c-1 & c & ab \\ \end{matrix}\, \right|=\left| \,\begin{matrix} a & a & bc \\ b & b & ca \\ c & c & ab \\ \end{matrix}\, \right|-\left| \,\begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \\ \end{matrix}\, \right|\] = \[-\left| \,\begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix}\, \right|=-\left| \,\begin{matrix} a & {{a}^{2}} & 1 \\ b-a & {{b}^{2}}-{{a}^{2}} & 0 \\ c-a & {{c}^{2}}-{{a}^{2}} & 0 \\ \end{matrix}\, \right|\] [By\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}};\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]] = \[-(a-b)\,(b-c)\,(c-a)\].
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