Question 101221
{{{(x^2-y^2)/(x^4-x^3)}}}÷{{{(x-y)/x^2}}}÷{{{(x^2+2xy+y^2)/(x+y)}}}
**the /= the fraction bar not division 
<pre><b>
We take care of the ÷ 's first in order left to right. So we put
parentheses arould the first two fractions so that we do that
division first.

({{{(x^2-y^2)/(x^4-x^3)}}} ÷ {{{(x-y)/x^2}}}) ÷ {{{(x^2+2xy+y^2)/(x+y)}}}

Within the parentheses we invert the second fraction and change
the division symbol to a multiplication symbol

({{{(x^2-y^2)/(x^4-x^3)}}} · {{{x^2/(x-y)}}}) ÷ {{{(x^2+2xy+y^2)/(x+y)}}}

Now we invert the third fraction and change the division symbol
to a multiplication:

({{{(x^2-y^2)/(x^4-x^3)}}} · {{{x^2/(x-y)}}}) · {{{(x+y)/(x^2+2xy+y^2)}}}

Now we don't need the parentheses around the first two any more:

{{{(x^2-y^2)/(x^4-x^3)}}} · {{{x^2/(x-y)}}} · {{{(x+y)/(x^2+2xy+y^2)}}}

Now we indicate the product of all the numerators over the
product of all the denominators:

{{{ ((x^2-y^2)(x^2)(x+y))   / ( (x^4-x^3)(x-y)(x^2+2xy+y^2) )}}}

Now we factor the {{{(x^2-y^2)}}} as {{{(x-y)(x+y)}}},
the {{{x^4-x^3}}} as {{{x^3(x-1)}}}, and
the {{{(x^2+2xy+y^2)}}} as {{{(x+y)(x+y)}}}

{{{ ((x-y)(x+y)(x^2)(x+y))   / ( x^3(x-1)(x-y)(x+y)(x+y) )}}}

We cancel the {{{(x-y)}}}in the top with the {{{(x-y)}}} in the bottom.
There are two {{{(x+y)}}} factors in the top that cancel
with the two {{{(x+y)}}} factors in the bottom.
The {{{(x^2)}}} in the top cancels into the {{{x^3}}} in the bottom,
leaving only an {{{x^1}}} or just {{{x}}} in the bottom. So we have only 
this left after canceling:

{{{ 1 / ( x(x-1) )}}}

Edwin</pre>