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If $\left|\begin{array}{lll}\mathrm{x} & \mathrm{y} & 0 \\ 0 & \mathrm{x} & \mathrm{y} \\ \mathrm{y} & 0 & \mathrm{x}\end{array}\right|=0$, then which one of the following is correct?
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The correct answer is:
$\frac{x}{y}$ is one of the cube roots of $-1$
$\left|\begin{array}{lll}x & y & 0 \\ 0 & x & y \\ y & 0 & x\end{array}\right|=0 \Rightarrow x\left(x^{2}-0\right)-y\left(0-y^{2}\right)+0=0$
$\Rightarrow x^{3}+y^{3}=0$
$\Rightarrow x^{3}=-y^{3}$
$\Rightarrow \frac{x^{3}}{y^{3}}=-1 \Rightarrow\left(\frac{x}{y}\right)^{3}=-1 \Rightarrow \frac{x}{y}=\sqrt{-1}$
$\Rightarrow x^{3}+y^{3}=0$
$\Rightarrow x^{3}=-y^{3}$
$\Rightarrow \frac{x^{3}}{y^{3}}=-1 \Rightarrow\left(\frac{x}{y}\right)^{3}=-1 \Rightarrow \frac{x}{y}=\sqrt{-1}$
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