A) 1
B) 0
C) \[(a-b)(b-c)(c-a)\]
D) \[(a+b)(b+c)(c+a)\]
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \\ \end{matrix}\, \right|\]=\[\left| \,\begin{matrix} 0 & 0 & 1 \\ c(b-a) & a(c-b) & ab \\ b-a & c+a & a+b \\ \end{matrix}\, \right|\] \[\{{{C}_{1}}\to {{C}_{1}}-{{C}_{2}},\,{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\}\] = \[(b-a)\,\,(c-b)\,\,\left| \,\begin{matrix} 0 & 0 & 1 \\ c & a & ab \\ 1 & 1 & a+b \\ \end{matrix}\, \right|\]= \[(b-a)\,(c-a)\,\,(c-a)\] \[=(a-b)\,\,(b-c)\,\,(c-a)\].
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