Question

6x4-35x3+62x2-35x+6=0

Answer 1

we know that $x=0$ doesn't satisfy the equation, therefore we can simply do this.
 
$6x^4-35x^3+62x^2-35x+6=0 \\ =>x^2(6(x^2+1/x^2)-35(x+1/x)+62)=0$

now take $x+1/x=y$ therefore you get 

$x^2(6(y^2-2)-35y+62)=0 \\ =>x^2(6y^2-35y+50)=0 \\ =>x^2(6y^2-15y-20y+50)=0 \\ =>x^2(2y-5)(3y-10)=0$

now put back $y=x+1/x$ and multiply each factor by $x$, you get

$(2x^2-5x+2)(3x^2-10x+3)=0 \\ =>(x-2)(2x-1)(x-3)(3x-1)=0 \\ =>x=2,3,1/2,1/3$.

Answer 2

first divide the complete equation by x².
then u will get,6x²-35x+62-35/x+6/x²=0
                   =6[x²+1/x²]-35[x+1/x]+62.----->eq.1
  now put x+1/x=a----->eq.2
then squaring on both sides,
x²+1/x²=a²-2.
now sub. the values in eq.1
then by solving u will get two values of a.
than sub. the two values in eq.2 and u will get four values for 'x'.