<s>[INST] 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
 [/INST] A nice trick:
As $x>0$, $$f(x)=\frac{\left(x+\cfrac1x\right)^6-\left(x^6+\cfrac1{x^6}\right)-2}{\left(x+\cfrac1x\right)^3+\left(x^3+\cfrac1{x^3}\right)}=\frac{6x^4+15x^2+18+\cfrac{15}{x^2}+\cfrac6{x^4}}{2x^3+3x+\cfrac3x+\cfrac2{x^3}}$$ giving $$f(x)=\frac{6x^8+15x^6+18x^4+15x^2+6}{2x^7+3x^5+3x^3+2x}=\frac{6x^8+9x^6+9x^4+6x^2}{2x^7+3x^5+3x^3+2x}+\frac{6x^6+9x^4+9x^2+6}{2x^7+3x^5+3x^3+2x}$$ which boils down to $$f(x)=3x+\frac3x.$$
</s>