Question 212871
Please help me.
{{{root(4,(5x^8y^3)/(27x^2))}}}
<pre><font size = 4 color = "indigo"><b>
Subtract the exponents of the {{{x}}}'s

{{{root(4,5x^6y^3/27)}}}

Write the {{{27}}}  as {{{3*3*3}}} and

{{{root(4,(5x^6y^3)/(3*3*3))}}}

Now take individual fourth roots of the
top and bottom.

{{{root(4,(5x^6y^3))/root(4,(3*3*3)))}}}


The denominator has only three factors of
3.  It needs one more 3, so it will be a 
perfect fourth power, and come out of the
fourth root radical.

So we multiply top and bottom of the fraction
by {{{red(3)}}}.

{{{root(4,( 5x^6y^3*red(3) )/(3*3*3*red(3)))}}}

Now that there are 4 factors of 3 in the
denominator, we can write the denominator as
{{{3^4}}} and simplify the numerator:

{{{root(4,(15x^6y^3 )/(3^4))}}} 

Now we take individual fourth roots of the
top and bottom:

{{{root(4,(15x^6y^3 ))/root(4,3^4))}}}

And since the bottom is the 4th root of
a 4th power we just take away the 4th root
radical and the 4th power.

{{{root(4,(15x^6y^3 ))/3}}}

The factor {{{x^6}}} is a case of an exponent 6
under a radical being as large or larger than, in
this case larger than, the index of the radical, 4.
That means it will simplify further.

So we write {{{x^6}}} as {{{red(x^4x^2)}}} so the {{{x^4}}}
will come out of the fourth root radical, like the 4
3's did above in the denominator.

{{{root(4,(15red(x^4x^2)y^3 ))/3}}}

Now you can take {{{x^4}}} out of the radical and
just have {{{x}}} in front of the radical on top:

{{{x*root(4,(15x^2y^3 ))/3}}}  

Edwin</pre>