Solve $6x^4-35x^3+62x^2-35x+6=0$

Given equation is a reciprocal equation of class one of even degree It is called a standard Reciprocal equation $6x^4-35x^3+62x^2-35x+6=0$ divide $x^2$ on both sides

$$\begin{gathered} \frac{6x^4}{x^2}35\frac{x^3}{x^2}+62\frac{x^2}{x^2}-\frac{35x}{x^2}+\frac{6}{x^2}=\frac{0}{x^2} \\ 6x^2-35x+62-\frac{35}{x}+\frac{6}{x^2}=0 \\ 6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0........\left(1\right) \\ put x+\frac{1}{x}=y \\ S.O.B.S \\ {\left(x+\frac{1}{x}\right)}^2=y^2; x^2+\frac{1}{x^2}+2=y^2 \\ {x}^2+\frac{1}{x^2}=y^2-2............\left(2\right)sub in \left(1\right) \\ 6\left(y^2-2\right)-35y+62=0 \\ 6y^2-12-35y+62=0 \\ 6y^2-35y+50=0 \\ 6y^2-20y-15y+50 \\ 3y-10 or 2y-5=0 \end{gathered}$$

$$\begin{gathered} Case-I Case-II \\ If 3y-10=0 If 2y-5=0 \\ 3\left(x+\frac{1}{x}\right)-10=0 2\left(x+\frac{1}{x}\right)-5=0 \\ \frac{3x^2+3-10x}{x}=0 \frac{2x^2+2-5x}{x}=0 \\ 3x^2-10x+3=0 2x^2-5x+2=0 \\ 3x^2-9x-x+3=0 2x^2 -4x-x+2 \\ \left(3x-1\right)\left(x-3\right)=0 \left(2x-1\right)\left(x-2\right)=0 \\ 3x-1=0, x-3=0 2x-1=0 or x-2=0 \\ x=1/3, x=3 2x=1 x=2 \\ x=1/3, 3 x=1/2 x=2 \\ \therefore x=1/2, 2 \\ The roots are 2, \frac{1}{2}, 3, \frac{1}{3} \end{gathered}$$