The first option is correct: we have
[tex]\left(k^{\frac{1}{8}}\right)^{-1} = \dfrac{1}{k^{\frac{1}{8}}} = \dfrac{1}{\sqrt[8]{k}},\quad \left(k^{-1}\right)^{\frac{1}{8}} = \left(\dfrac{1}{k}\right)^{\frac{1}{8}} = \dfrac{1}{\sqrt[8]{k}}[/tex]
The second option is also correct, because it simply applies the definition
[tex]k^{\frac{1}{n}} = \sqrt[n]{k}[/tex]
The third option is also correct, because it applies the rule
[tex](a^b)^c = a^{bc}[/tex]
