Question 393123
<pre><font face = "batangche" color = "indigo" size = 4><b>

{{{(7x)/root(3,4xy^5)}}}

Break the denominator down into prime factors:

{{{(7x)/root(3,2*2*x*y*y*y*y*y)}}}

We now must decide what factors the denominator would need in order 
for it to become a perfect cube.  We need three or a multiple of 
three of each factor:

We have 2*2, which is two factors, so we need one more factor of 2
so we'll have three factors of 2.

We have x, which is one factor, so we need two more factors of x
so we'll have three factors of x.

We have y*y*y*y*y, which is five factors, so we need one more factor of y
so we'll have a multiple of three, that is, six factors of y.

So we create another cube root which has what we need in it, one
factor of 2, two factors of x and one factor of y.  That is, we
create this cube root:

{{{root(3,2*x*x*y)}}}

Then we place it over itself so that the value will be 1.

{{{root(3,2*x*x*y)/root(3,2*x*x*y)}}}

And we can now multiply it by the original expression without
changing its value since we are actually only multiplying by 1.

{{{(7x)/root(3,2*2*x*y*y*y*y*y)}}}{{{""*""}}}{{{root(3,2*x*x*y)/root(3,2*x*x*y)}}}

Now we multiply under the radicals on the bottom:

{{{(7x*root(3,2*x*x*y))/root(3,2*2*2*x*x*x*y*y*y*y*y*y)}}}

Now we group the like factors in the bottom into groups of three:

{{{(7x*root(3,2*x*x*y))/root(3,(2*2*2)*(x*x*x)*(y*y*y)*(y*y*y))}}}

Now we write each group of three as the cube of a single factor:

{{{(7x*root(3,2*x*x*y))/root(3,2^3*x^3*y^3*y^3)}}}

Now take individual cube roots:

{{{(7x*root(3,2*x*x*y))/(2*x*y*y)}}}

Write the {{{x*x}}} in the top as {{{x^2}}} and the {{{y*y}}} in
the bottom as {{{y^2}}}

{{{(7x*root(3,2x^2y))/(2xy^2)}}}

You can perhaps figure out a shorter way than this, but this method
will always work.  Do them this way and I think you will figure out 
a shortcut after doing a few this longer but easier to follow way.

Edwin</pre>