Since $ xyz = 1, $ the numerator is \begin{align*} 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{align*}Similarly, the denominator is \begin{align*} 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{align*}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)} $.