Convert to Radical Form (64n^12)^(-1/6)

${(64 n^{12})}^{- \frac{1}{6}}$

Step 1

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

$\frac{1}{{(64 n^{12})}^{\frac{1}{6}}}$

Step 2

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{1}{\sqrt[6]{{(64 n^{12})}^{1}}}$

Step 3

Anything raised to $1$ is the base itself.

$\frac{1}{\sqrt[6]{64 n^{12}}}$