A) \[abc=1\]
B) \[abc=-1\]
C) \[a+b+c=0\]
D) \[a+b+c=\pm 1\]
Correct Answer: A
Solution :
\[\left| \begin{matrix} 1-{{a}^{3}} & {{a}^{2}} & a \\ 1-{{b}^{3}} & {{b}^{2}} & b \\ 1-{{c}^{3}} & {{c}^{2}} & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & {{a}^{2}} & a \\ 1 & {{b}^{2}} & b \\ 1 & {{c}^{2}} & c \\ \end{matrix} \right|-abc\left| \begin{matrix} {{a}^{2}} & a & 1 \\ {{b}^{2}} & b & 1 \\ {{c}^{2}} & c & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} {{a}^{2}} & a & 1 \\ {{b}^{2}} & b & 1 \\ {{c}^{2}} & c & 1 \\ \end{matrix} \right|[1-abc]=0\] \[\Rightarrow \] \[\left| \begin{matrix} {{a}^{2}} & a & 1 \\ {{b}^{2}} & b & 1 \\ {{c}^{2}} & c & 1 \\ \end{matrix} \right|\ne 0\] \[\therefore \] \[1-abc=0\] \[\Rightarrow \] \[abc=1\]
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