A) \[1+abc=0\]
B) \[a+b+c+1=0\]
C) \[(a-b)(b-c)(c-a)=0\]
D) None of these
Correct Answer: A
Solution :
Splitting the determinant into two determinants, we get \[\Delta =\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|+abc\,\left| \,\begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix}\, \right|\,=0\] = \[(1+abc)\,[(a-b)\,(b-c)\,(c-a)]=0\] Because a, b, c are different, the second factor cannot be zero. Hence, option (a), \[1+abc=\]0, is correct.
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