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?