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