A) 0
B) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
C) \[3abc\]
D) \[{{(a+b+c)}^{3}}\]
Correct Answer: A
Solution :
\[\left| \,\begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|=\left| \,\begin{matrix} 0 & a-b & (a-b)\,(a+b+c) \\ 0 & b-c & (b-c)\,\,(a+b+c) \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|\]by \[\left\{ \begin{align} & {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\ & {{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\ \end{align} \right.\] = \[(a-b)\,(b-c)\,\left| \,\begin{matrix} 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix}\, \right|=0\], \[\{\because \,\,{{R}_{1}}\equiv {{R}_{2}}\}\].
You need to login to perform this action.
You will be redirected in
3 sec
