30. Let $x+\frac{1}{x}=\sqrt{2}$. Find the value of $x^{8}+\frac{1}{x^{8}}$.
(ans. 2 .

Let $x+\frac{1}{x}=2 \cos \theta \Rightarrow x^{2}-2 \cos \theta \cdot x+1=0 \Rightarrow x=\cos \theta \pm i \sin \theta \Rightarrow x^{n}=$ $\cos n \theta \pm i \sin n \theta$ and $x^{-n}=\cos n \theta \mp i \sin n \theta$. Thus, $x^{n}+\frac{1}{x^{n}}=2 \cos n \theta$ Let $\sqrt{2}=2 \cos (\pi / 4)$ and $n=8$ so that $x^{8}+\frac{1}{x^{8}}=2 \cos 8(\pi / 4)=2$.)