jmc

Let $$f(x)=\frac{{\left(x+\frac{1}{x}\right)}^6-\left(x^6+\frac{1}{x^6}\right)-2}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)}.$$We can write $$f(x)=\frac{{\left(x+\frac{1}{x}\right)}^6-\left(x^6+2+\frac{1}{x^6}\right)}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)}=\frac{{\left(x+\frac{1}{x}\right)}^6-{\left(x^3+\frac{1}{x^3}\right)}^2}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)}.$$By difference of squares, $$\begin{array}{rl}f(x)& =\frac{\left[{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)\right]\left[{\left(x+\frac{1}{x}\right)}^3-\left(x^3+\frac{1}{x^3}\right)\right]}{{\left(x+\frac{1}{x}\right)}^3+\left(x^3+\frac{1}{x^3}\right)}\\ & ={\left(x+\frac{1}{x}\right)}^3-\left(x^3+\frac{1}{x^3}\right)\\ & =x^3+3x+\frac{3}{x}+\frac{1}{x^3}-x^3-\frac{1}{x^3}\\ & =3x+\frac{3}{x}\\ & =3\left(x+\frac{1}{x}\right).\end{array}$$By AM-GM, $x+\frac{1}{x}\ge 2,$ so $$f(x)=3\left(x+\frac{1}{x}\right)\ge 6.$$Equality occurs at $x=1,$ so the minimum value of $f(x)$ for $x>0$ is $\boxed{6}.$