The real x satisfying the equation x3+1x3+x2+1x2−6x+1x−7=0 is/are

Let x+1x=tThen given equation ist3−3t+t2−2−6(t)−7=1 or     t3+t2−9t−9=0∴    (t+1)t2−9=0∴    t=−1,3,−3If x+1x=−1, then x2+x+1=0. This equation has no real roots.If x+1x=−3, then x2+3x+1=0. Roots are −3±52.If x+1x=3, then x2−3x+1=0. Roots are 3±52.