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