If ${x}^2+\frac{1}{{x}^2}=2$ then the value of ${x}^{256}+\frac{1}{{x}^{256}}$

Given \(x^2+\dfrac{1}{x^2}=2\). Let \(t=x+\dfrac{1}{x}\). Then \(x^2+\dfrac{1}{x^2}=t^2-2\), so \(t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2\).

From \(x+\dfrac{1}{x}=2 \Rightarrow x=1\) and from \(x+\dfrac{1}{x}=-2 \Rightarrow x=-1\).

Hence \(x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}\).

Final Answer: \(2\).