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The number of real roots of the equation $\left(x+\frac{1}{x}\right)^{3}+x+\frac{1}{x}=0$ is
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$\begin{array}{l}
\left(x+\frac{1}{x}\right)^{3}+\left(x+\frac{1}{x}\right)=0 \\
\Rightarrow\left(x+\frac{1}{x}\right)+\left[\left(x+\frac{1}{x}\right)^{2}+1\right]=0 \\
\Rightarrow x+\frac{1}{x}=0 \Rightarrow x^{2}+1=0 \Rightarrow x=\pm i
\end{array}$
Thus, the given equation has no real roots.
\left(x+\frac{1}{x}\right)^{3}+\left(x+\frac{1}{x}\right)=0 \\
\Rightarrow\left(x+\frac{1}{x}\right)+\left[\left(x+\frac{1}{x}\right)^{2}+1\right]=0 \\
\Rightarrow x+\frac{1}{x}=0 \Rightarrow x^{2}+1=0 \Rightarrow x=\pm i
\end{array}$
Thus, the given equation has no real roots.
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