Solve $6x^6-25x^5+31x^4-31x^2+25x-6=0$

Given equation is a reciprocal equation of class two and of even degree then 1 and –1 are roots 

$\therefore$ $6x^6-25x^5+31x^4-31x^2+25x-6=0$

$=(x^2-1)(6x^4-25x^3+37x^2-25x+6)=0$

Now $6x^4-25x^3+37x^2-25x+6=0.......\left(1\right)$

A reciprocal equation of class one of even degree is callel a standard reciprocal equation divided by $x^2$ an both sides of equation (1) 

$$\begin{gathered} \frac{6x^4}{x^2}-\frac{25x^3}{x^2}+\frac{37x^2}{x^2}-\frac{25x}{x^2}+\frac{6}{x^2}=\frac{0}{x^2} \\ 6x^2-25x+37-\frac{25}{x}+\frac{6}{x^2}=0 \\ 6\left(x^2+\frac{1}{x^2}\right)-25\left[x+\frac{1}{x}\right]+37=0......\left(2\right) \\ Put x+\frac{1}{x}=y \\ S.O.BS \\ {\left(x+\frac{1}{x}\right)}^2=y^2; x^2+\frac{1}{x^2}+2=y^2 \\ {x}^2+\frac{1}{x^2}=y^2-2 sub in \left(2\right) \\ 6\left(y^2-2\right)-25y+25=0 \\ \left(2y-5\right)\left(3y-5\right)=0 \\ 2y-5=0 or 3y-5=0 \end{gathered}$$

$$\begin{gathered} Case I Case II \\ If 2y-5=0 If 3y-5=0 \\ 2\left(x+\frac{1}{x}\right)-5=0 3\left(x+\frac{1}{x}\right)-5=0 \\ 2\left(\frac{x^2+1}{x}\right)-5=0 3\left(\frac{x^2+1}{x}\right)-5=0 \\ \frac{2x^2+2-5x}{x}=0 \frac{3x^2+3-5x}{x}=0 \\ 2x^2-5x+2=0 3x^2-5x+3=0 \\ \left(x-2\right)\left(2x-1\right)=0 x=-\frac{b\pm \sqrt{b^2-4ac}}{2a} \\ x-2=0, 2x-1=0 =\frac{5\pm \sqrt{25-36}}{6} \\ x=2, x=\frac{1}{2} =\frac{5\pm i\sqrt{11}}{6} \\ \because The roots are \pm 1, 2, \frac{1}{2}, \frac{5\pm i\sqrt{11}}{6} \end{gathered}$$