jmc

Since $xyz=1,$ the numerator is $$\begin{array}{rl}{x}^3+y^3+z^3-x^{-3}-y^{-3}-z^{-3}& =x^3+y^3+z^3-y^3{z}^3-x^3{z}^3-x^3{y}^3\\ & =x^3{y}^3{z}^3-x^3{y}^3-x^3{z}^3-y^3{z}^3+x^3+y^3+z^3-1\\ & =(x^3-1)(y^3-1)(z^3-1).\end{array}$$Similarly, the denominator is $$\begin{array}{rl}x+y+z-x^{-1}-y^{-1}-z^{-1}& =x+y+z-xy-xz-yz\\ & =xyz-xy-xz-yz+x+y+z-1\\ & =(x-1)(y-1)(z-1).\end{array}$$Therefore, the given expression is equal to $$\frac{(x^3-1)(y^3-1)(z^3-1)}{(x-1)(y-1)(z-1)}=(x^2+x+1)(y^2+y+1)(z^2+z+1).$$By AM-GM, $$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge (3x)(3y)(3z)=27xyz=27.$$The only way to get equality is if $x=y=z=1.$ However, this cannot occur, as this would make the given expression undefined.

If we set $y=x,$ then the given expression is equal to $$(x^2+x+1)(x^2+x+1)\left(\frac{1}{x^4}+\frac{1}{x^2}+1\right).$$Letting $x$ approach 1 from above, and letting $x$ approach $\infty ,$ we see that the given expression can attain any value in $\boxed{(27,\infty )}.$