Question 64849
<pre><font size = 5><b>{{{(x^2a/z^3 - a^2)/(x/z^2 - 3/z^3)}}}
Put the a² over 1<sup>1</sup> so that everything on top 
and bottom will be a fraction.  I used 1<sup>1</sup> instead
of just 1 because it "stacks up" better using the forum's
notation system.  You can just use 1 where I have 1<sup>1</sup>.
{{{(x^2a/z^3 - a^2/1^1)/(x/z^2 - 3/z^3)}}}
Look at all the denominators on the top and the bottom.
Their LCD is z³. so multiply by {{{z^3/z^3}}}, written as{{{ ((z^3/1^1))/((z^3/1^1))}}}


{{{ ((z^3/1^1))/((z^3/1^1))}}}·{{{((x^2a/z^3 - a^2/1^1))/((x/z^2 - 3/z^3))}}}
Now distribute:
{{{ ((z^3/1^1)(x^2a/z^3) - (z^3/1^1)(a^2/1^1))/((z^3/1^1)(x/z^2)-(z^3/1^1)(3/z^3))}}}     
In the upper left term you can cancel the z³'s, 
     leaving x²a.
The upper right term becomes just -z³a².
In the lower left term, subtract the exponents 
     of the z's and get simply zx.
In the lower right term you can cancel the z³'s,
     leaving -3.

So you end up with:
{{{(x^2a - z^3a^2)/(zx-3)}}}
Now if you like you can factor out "a" on top:
{{{a(x^2 - z^3a)/(zx-3)}}}
but that step isn't necessary.

Edwin</pre>