A) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\]
B) \[{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\]
C) 0
D) \[-{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\]
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} 0 & {{b}^{3}}-{{a}^{3}} & {{c}^{3}}-{{a}^{3}} \\ {{a}^{3}}-{{b}^{3}} & 0 & {{c}^{3}}-{{b}^{3}} \\ {{a}^{3}}-{{c}^{3}} & {{b}^{3}}-{{c}^{3}} & 0 \\ \end{matrix}\, \right|\] \[({{b}^{3}}-{{a}^{3}})({{c}^{3}}-{{a}^{3}})\left| \,\begin{matrix} 0 & 1 & 1 \\ {{a}^{3}}-{{b}^{3}} & 1 & 1 \\ {{a}^{3}}-{{c}^{3}} & 1 & 1 \\ \end{matrix}\, \right|=0\] \[[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}]\] and then taking out common \[({{b}^{2}}-{{a}^{3}})\]from IInd column and ( \[{{c}^{3}}-{{a}^{3}}\]) from IIIrd column].
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