smc

Let $a={\left(x+\frac{1}{x}\right)}^3$ and $b=x^3+\frac{1}{x^3}$. Then $$\begin{array}{rl}f(x)& =\frac{{\left(x+\frac{1}{x}\right)}^6-(x^6+\frac{1}{x^6})-2}{{\left(x+\frac{1}{x}\right)}^3+(x^3+\frac{1}{x^3})}\\ & =\frac{{\left(x+\frac{1}{x}\right)}^6-(x^6+2+\frac{1}{x^6})}{{\left(x+\frac{1}{x}\right)}^3+(x^3+\frac{1}{x^3})}\\ & =\frac{{\left(x+\frac{1}{x}\right)}^6-(x^3+\frac{1}{x^3}{)}^2}{{\left(x+\frac{1}{x}\right)}^3+(x^3+\frac{1}{x^3})}\\ & =\frac{a^2-b^2}{a+b}.\end{array}$$ By difference of squares, $$\begin{array}{rl}f(x)& =\frac{(a-b)(a+b)}{a+b}\\ & =a-b\\ & ={\left(x+\frac{1}{x}\right)}^3-\left(x^3+\frac{1}{x^3}\right)\\ & =\left(x^3+3x+\frac{3}{x}+\frac{1}{x^3}\right)-\left(x^3+\frac{1}{x^3}\right)\\ & =3x+\frac{3}{x}\\ & =3\left(x+\frac{1}{x}\right).\end{array}$$ By the AM-GM inequality, $$x+\frac{1}{x}\ge 2,$$ so $f(x)\ge 6$. Furthermore, when $x=1$, $f(1)=6$, so the minimum value of $f(x)$ is $\boxed{6}$.