A) \[3abc+{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\]
B) \[3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\]
C) \[abc-{{a}^{3}}+{{b}^{3}}+{{c}^{3}}\]
D) \[abc+{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\]
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix}\, \right|=\left| \,\begin{matrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \\ \end{matrix}\, \right|\], \[({{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}})\] = \[(a+b+c)\] \[k=1,\,-1.\] =\[3abc-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}\], (After simplification).You need to login to perform this action.
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