so remember some exponentional laws
\(x^{m/n} = \sqrt[n]{x^m}\)
\(\frac{x^n}{x^m}=x^{n-m}\) and
\(x^{-n}= \frac{1}{x^{n}}\)
so
4 to the 6th root=4^(1/6)
4 to the 3rd root=4^(1/3)
so
\(\frac{4^{1/6}}{4^{1/3}} =4^{(1/6)-(1/3)}=4^{(1/6)-(2/6)}=4^{-1/3}\)
and \(4^{-1/3}= \frac{1}{4^{1/3}}= \frac{1}{ \sqrt[3]{4} }\)
the answer is \(\frac{1}{ \sqrt[3]{4} }\)
