Question 123577

Solution by Edwin:</font>

<pre><b>
{{{sqrt((54w^16)/(150z^4))}}}

Break everything into prime factors

{{{54 = 2*3*3*3}}}

{{{w^16 = wwwwwwwwwwwwwwww}}}

{{{150 = 2*3*5*5}}}

{{{z^4 = zzzz}}}

So 

{{{sqrt((54w^16)/(150z^4))}}}

becomes

{{{sqrt((2*3*3*3wwwwwwwwwwwwwwww)/(2*3*5*5zzzz))}}}

Put parentheses arond every pair of like factors which are multiplied:

{{{sqrt((2(3*3)3(ww)(ww)(ww)(ww)(ww)(ww)(ww)(ww))/(2*3(5*5)(zz)(zz)))}}}

Every pair of multiplied like factors is the square of that factor,
so write each pair as a square:

{{{sqrt(  ( 2(3^2)3w^2w^2w^2w^2w^2w^2w^2w^2 )  / (2*3 (5^2z^2z^2) ) ) }}}

All the squares on top come out in front of the radical without an exponent
as a numerator.  All those on the bottom come out likewise but are in the
denominator in front.  Every factor that did not pair up stays under the
radical:

{{{( (3wwwwwwww)/(5zz) ) sqrt((2*3)/(2*3))}}}

Now write {{{w^8}}} for {{{wwwwwwww}}} and {{{z^2}}} for {{{zz}}}
and what's under the radical cancels out and leaves just {{{1}}}

{{{( (3w^8)/(5z^2) ) sqrt(cross(2*3)/cross(2*3))}}}

or

{{{( (3w^8)/(5z^2) )1}}}

or

{{{ (3w^8)/(5z^2) }}}

Edwin</pre>