Question 981109
Same idea as with square root in the denominator; but you just have cubed root, meaning you need this
as a factor shown three times to reach the expression inside.


(sqrt( 2 xy^2))/(cube root of( 8 xy^3))


{{{(sqrt(2xy^2))/(root(3,8 xy^3))}}}


{{{((sqrt(2xy^2))/(root(3,8 xy^3)))((root(3,8 xy^3))/(root(3,8 xy^3)))((root(3,8 xy^3))/(root(3,8 xy^3)))}}}


{{{(sqrt(2xy^2)root(3,8xy^3)root(3,8xy^3))/(8xy^3)}}}


This may not be the final form that you want, but this is now an equivalent expression
and the denominator is now rationalized.