A) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
B) \[3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\]
C) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-{{a}^{2}}b-{{b}^{2}}c-{{c}^{2}}a\]
D) \[3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\]
Correct Answer: B
Solution :
\[\Delta =\left| \,\begin{matrix} 2(a+b+c) & 0 & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \\ \end{matrix}\, \right|\] by \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}\] \[\Delta =(a+b+c)\,.\,\left| \,\begin{matrix} 2 & 0 & 1 \\ c+a & b-c & b \\ a+b & c-a & c \\ \end{matrix}\, \right|\] On expanding, \[-(a+b+c)\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] = \[=a_{1}^{2}({{b}_{2}}{{c}_{3}}-{{b}_{3}}{{c}_{2}})+{{a}_{1}}{{b}_{1}}(-{{c}_{3}}{{a}_{2}}+{{a}_{3}}{{c}_{2}})\]. Trick: Put \[a=1,\,b=2,\,c=3\] and check it.
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