Simplify the Radical Expression (k^(1/8))^-1

${(k^{\frac{1}{8}})}^{- 1}$

Multiply the exponents in ${(k^{\frac{1}{8}})}^{- 1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$k^{\frac{1}{8} \cdot - 1}$

Combine $\frac{1}{8}$ and $- 1$ .

$k^{\frac{- 1}{8}}$

Move the negative in front of the fraction.

$k^{- \frac{1}{8}}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{k^{\frac{1}{8}}}$