A) \[z=xy\]
B) \[z=\frac{1}{xy}\]
C) \[z=-\frac{1}{xy}\]
D) None of these
Correct Answer: C
Solution :
Given that, \[\left| \begin{matrix} x & {{x}^{2}} & 1+{{x}^{3}} \\ y & {{y}^{2}} & 1+{{y}^{3}} \\ z & {{z}^{2}} & 1+{{z}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} x & {{x}^{2}} & 1 \\ y & {{y}^{2}} & 1 \\ z & {{z}^{2}} & 1 \\ \end{matrix} \right|+xyz\left| \begin{matrix} 1 & x & {{x}^{2}} \\ 1 & y & {{y}^{2}} \\ 1 & z & {{z}^{2}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(1+xyz)=\left| \begin{matrix} 1 & x & {{x}^{2}} \\ 1 & y & {{y}^{2}} \\ 1 & z & {{z}^{2}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[xyz+1=0\Rightarrow z=-\frac{1}{xy}\]
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