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