$\begin{array}{l}GivenEquation{x}^5-5x^4+9x^3-9x^2+5x-1=0is\\ reciprocalEquation\therefore x=1isroot\end{array}$

$\begin{array}{l}\therefore x^4-4x^3+5x^2-4x+1=0\\ \Rightarrow x^2-4x+5-\frac{4}{x}+\frac{1}{x^2}=0\\ \Rightarrow (x^2+\frac{1}{x^2})-4(x+\frac{1}{x})+5=0\\ Letx+\frac{1}{x}=y\Rightarrow x^2+\frac{1}{x^2}=y^2-2\\ \therefore y^2-2-4y+5=0\\ \Rightarrow y^2-4y+3=0\\ \Rightarrow y^2-3y-y+3=0\\ \Rightarrow (y-1)(y-3)=0\end{array}$

$\begin{array}{l}y=1,y=3\\ Ify=1thenx+\frac{1}{x}=1\\ \Rightarrow x^2-x+1=0\\ x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm i\sqrt{3}}{2}\\ Ify=3\Rightarrow x+\frac{1}{x}=3\\ \Rightarrow x^2-3x+1=0\\ x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\\ Differenceofroots=\frac{3+\sqrt{5}}{2}-\left(\frac{3-\sqrt{5}}{2}\right)=\sqrt{5}\end{array}$