If 𝒙 = 𝟏 − √𝟐, find the value of \left(x+\frac{1}{x}\right)^{3}
\begin{aligned} &\text { Qwen that } x=1-2\\ &\frac{1}{2}-\frac{1}{1-\sqrt{2}} \cdot \frac{1+\sqrt{2}}{1+\sqrt{2}}\\ &=\frac{1}{1}=\frac{t+r}{1-1+2 !}\\ &[\sin c=\operatorname{san} b x]+t\}=a-t^{2}\\ &=\frac{1}{x}-\frac{1+\sqrt{2}}{1-x}\\ &=\frac{1}{x}=\frac{1+\sqrt{2}}{-1}\\ &\left.\Rightarrow \frac{1}{x}=-\frac{1}{1} 1+\sqrt{2}\right\} \quad(1)\\ &\Rightarrow\left(x-\frac{t}{x}\right)=1-22+t+\sqrt{2}\\ &\Rightarrow\left[x-\frac{1}{2}\right]-2\\ &\Rightarrow\left(x-\frac{1}{x}\right)^{2} \cdot 2^{x}\\ &-\left(x-\frac{1}{x}\right)^{2}-8 \end{aligned}
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