Solve $x^5-5x^4+9x^3-9x^2+5x-1=0$

Given equation is a reciprocal equation of odd degree and of class two 

$\therefore$1 is a root of the given equation By synthetic division 

By synthetic division

Now $\left(x-1\right)\left(x^4-4x^3+5x^2-4x+1\right)=0$

$Now x^4-4x^3+5x^2-4x+1=0$is a standard reciprocal equation divided by $x^2$ on both sides 

$$\begin{gathered} \frac{x^4}{x^2}-\frac{4x^3}{x^2}+\frac{5x^2}{x^2}-\frac{4x}{x^2}+\frac{1}{x^2}=\frac{0}{x^2} \\ {x}^2-4x+5-\frac{4}{x}+\frac{1}{x^2}=0 \\ \left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+5=0 \\ Put x+\frac{1}{x}=y \\ S.O.B.S {\left(x+\frac{1}{x}\right)}^2=y^2 ; x^2+\frac{1}{x^2}+2=y^2 \\ {x}^2+\frac{1}{x^2}=y^2-2 \to sub in \left(1\right) \\ \left(y^2-2\right)-4y+5=0 ; y^2-4y+3=0 \\ \left(y-1\right)\left(y-3\right)=0 ; y-1=0 or y-3=0 \end{gathered}$$

$$\begin{gathered} Case-I Case-II \\ If y-1=0 If y-3=0 \\ x+\frac{1}{x}-1=0 x+\frac{1}{x}-3=0 \\ \frac{x^2+1-x}{x}=0 \frac{x^2+1-3x}{x}=0 \\ {x}^2-x+1=0 x^2-3x+1=0 \\ x=\frac{1\pm \sqrt{1-4}}{2} x=\frac{3\pm \sqrt{9-4}}{2} \\ =\frac{1\pm i\sqrt{3}}{2} =\frac{3\pm \sqrt{5}}{2} \\ The roots of given equation are 1, \frac{1\pm i\sqrt{3}}{2}, \frac{3\pm \sqrt{5}}{2} \end{gathered}$$