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