Question 434542
<pre>
{{{root(3,3x^2y^2)/root(3,48x^6y)}}}

Write as a single cube root:

{{{root(3,(3x^2y^2)/(48x^6y))}}}


Divide top and bottom by 3, which gives 16 on the bottom
Subtract the exponents of x, getting x<sup>4</sup> on the bottom.
Subtract the exponents of y, getting y on the top:

{{{root(3,y/(16x^4))}}}

rewrite the 16 as 2<sup>4</sup>

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

The idea is to get the denominator so that all the exponents will
be divisible by 3, the index of the root

To make 2 have an exponent divisible by 3, we need to multiply it
by 2<sup>2</sup> so it will have an exponent of 6

To make x have an exponent divisible by 3, we need to multiply it
by x<sup>2</sup> so it will have an exponent of 6.
So we multiply under the radical by {{{red((2x^2)/(2x^2))}}}
{{{root(3,expr(y/(2^4x^4))*red(expr((2^2x^2)/(2^2x^2)))) }}}

or

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

Separate into the quotient of two cube roots again:

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

Get rid of the cube root on the bottom by dividing
each exponent by the index 3 of the root:

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

Changing 2<sup>2</sup>'s to 4's

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


Edwin</pre>