A) \[a+b+c=0\]
B) \[abc=1\]
C) \[a+b+c=1\]
D) \[ab+bc+ca=0\]
Correct Answer: B
Solution :
\[\left| \,\begin{matrix} a & {{a}^{2}} & {{a}^{3}}-1 \\ b & {{b}^{2}} & {{b}^{3}}-1 \\ c & {{c}^{2}} & {{c}^{3}}-1 \\ \end{matrix}\, \right|=0\]Þ \[\left| \,\begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix}\, \right|-\left| \,\begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix}\, \right|=0\] Þ \[abc\,\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|-\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|=0\] Þ \[(abc-1)\,\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|=0\] Since \[a,b,c\] are different, so \[\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|\ne 0\] Hence \[abc-1=0\]i.e., \[abc=1\].
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