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