Determine Minimum Value. 
Find the minimum value of
  $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
  for $x>0$.

When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks
I think differentiation would be really complicated

Hint-$$\frac{(x+\frac1x)^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})} = \frac{(x+\frac{1}{x})^6-(x^3+\frac{1}{x^3})^2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}=(x+\frac{1}{x})^3-(x^3+\frac{1}{x^3})=3(x+1/x)$$
