Question 460172
{{{root(3,(2x^4y^4)/(9x))}}}
 <pre><font face = "consolas" color = "indigo" size = 4><b>
First we cancel the x in the bottom into the x&#8308;
 in the top getting x³ in the top

{{{root(3,(2x^3y^4)/(9))}}}

Next we break the cube root of a quotient 
into the quotient of two cube roots:

{{{root(3,2x^3y^4)/root(3,9)}}}

Next we want to get rid of the cube root on
the bottom. It contains 9 = 3*3.  Let's write
9 as 3*3

{{{root(3,2x^3y^4)/root(3,3*3)}}}

We can make that 3*3 into the perfect cube 3*3*3 
or 3³ by multiplying it by 3, so we multiply the 
entire fraction by {{{root(3,3)/root(3,3)}}}

{{{expr(root(3,2x^3y^4)/root(3,3*3))expr(root(3,3)/root(3,3))}}}

Now we multiply numerators and denominators by 
multiplying under radicals:

{{{root(3,2*3x^3y^4)/root(3,3*3*3)}}}
  
{{{root(3,6x^3y^4)/root(3,3^3)}}}

Now the denominator becomes just 3 and we write
the y&#8308;in the numerator as y³y:

{{{root(3,6x^3y^3y)/3}}}

Finally we take the x³y³ out of the radical in 
the numerator and put xy in front:

{{{(xy*root(3,6y))/3}}}

Edwin</pre>