A) \[a+b+c\]
B) 0
C) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
D) \[{{a}^{2}}-{{b}^{2}}+{{c}^{2}}\]
Correct Answer: A
Solution :
\[\left[ \begin{matrix} a & {{a}^{3}} & {{a}^{4}}-1 \\ b & {{b}^{3}} & {{b}^{4}}-1 \\ c & {{c}^{3}} & {{c}^{4}}-1 \\ \end{matrix} \right]\,=0\] or \[\left| \,\begin{matrix} a & {{a}^{3}} & {{a}^{4}} \\ b & {{b}^{3}} & {{b}^{4}} \\ c & {{c}^{3}} & {{c}^{4}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} a & {{a}^{3}} & -1 \\ b & {{b}^{3}} & -1 \\ c & {{c}^{3}} & -1 \\ \end{matrix}\, \right|=0\] or \[abc\text{ }\left| \,\begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\ 1 & {{b}^{2}} & {{b}^{3}} \\ 1 & {{c}^{2}} & {{c}^{3}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} a & {{a}^{3}} & -1 \\ a-b & {{a}^{3}}-{{b}^{3}} & 0 \\ a-c & {{a}^{3}}-{{c}^{3}} & 0 \\ \end{matrix}\, \right|\,=0\] or \[abc\text{ }\left| \,\begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\ 0 & {{a}^{2}}-{{b}^{2}} & {{a}^{3}}-{{b}^{3}} \\ 0 & {{a}^{2}}-{{c}^{2}} & {{a}^{3}}-{{c}^{3}} \\ \end{matrix}\, \right|+\left| \,\begin{matrix} a & {{a}^{3}} & -1 \\ a-b & {{a}^{3}}-{{b}^{3}} & 0 \\ (a-c) & ({{a}^{3}}-{{c}^{3}}) & 0 \\ \end{matrix}\, \right|\,=0\] or \[(abc)\,(a-b)\,(a-c)\,\left| \,\begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\ 0 & a+b & {{a}^{2}}+{{b}^{2}}+ab \\ 0 & a+c & {{a}^{2}}+{{c}^{2}}+ac \\ \end{matrix}\, \right|\,+\]\[(a-b)\,(a-c)\,\left| \,\begin{matrix} a & {{a}^{3}} & -1 \\ 1 & {{a}^{2}}+{{b}^{2}}+ab & 0 \\ 1 & {{a}^{2}}+{{c}^{2}}+ac & 0 \\ \end{matrix}\, \right|\] or \[(a-b)\,(a-c)\,[(abc)[(a+b)\,({{a}^{2}}+{{c}^{2}}+ac)-\]\[(a+c)({{a}^{2}}+{{b}^{2}}+ab)]]+(-1)\,(a-b)\,(a-c)\]\[[{{a}^{2}}+{{c}^{2}}+ac-{{a}^{2}}-{{b}^{2}}-ab]=0\] = \[(abc)\,[(a-b)\,(a-c)\,(c-b)(ac+ab+bc)]\]\[+(-1)(a-b)(a-c)(c-b)(a+b+c)=0\] \[\Rightarrow \] \[(abc)\,(ac+ab+bc)=a+b+c\].
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