By using the substitution y=x+1/x, find all the real roots of the equation (x^3) -4x+6-(4/x)+(1/x^3)
y=x+1/x……………………(1)
y^{2}=x^{2}+(1/x^2)+2……………...(2)
x^{2}+(1/x^2)=y^{2}-2
x^{3}-4x+6-4/x +(1/x^2) =0
(x^{2}+1/x^2 )(x+1/x)-(x+1/x)- 4(x+1/x)+6=0
From(1)and(2), (y^{2}-2)(y)-y-4y+6=0
y^{3}-7y+6=0
....
what should I do now?