Question 207462
<pre><font size = 4 color = "indigo"><b>{{{(3x+3y+3)/(9x)}}}÷{{{(x^2+2xy+y^2-1)/(x^4+x^2)}}}

Invert the second fraction and change the 
division to multiplication:

{{{((3x+3y+3)/(9x))((x^4+x^2)/(x^2+2xy+y^2-1))}}}

Factor {{{3x+3y+3}}} by factoring out {{{3}}} and getting {{{3(x+y+1)}}}

Factor {{{x^4+x^2}}} by factoring out {{{x^2}}} and getting {{{x^2(x^2+1)}}}

Factor {{{x^2+2xy+y^2-1}}} 
Factor the first three terms as 
{{{(x+y)^2-1}}}
and then factor that as the difference of two squares:
{{{((x+y)-1)((x+y)+1)}}}
or
{{{(x+y-1)(x+y+1)}}}

Now replace each by its factored form:

{{{((3(x+y+1))/(9x))((x^2(x^2+1))/((x+y-1)(x+y+1)))}}}

Cancel the 3 into the 9.

{{{((cross(3)(x+y+1))/(cross(9)[3]x))((x^2(x^2+1))/((x+y-1)(x+y+1)))}}}

The x in the bottom cancels out the square in the top:

{{{((cross(3)(x+y+1))/(cross(9)[3]cross(x)))((x^cross(2)(x^2+1))/((x+y-1)(x+y+1)))}}}

Cancel the {{{(x+y+1)}}}'s:

{{{((cross(3)(cross(x+y+1)))/(cross(9)[3]cross(x)))((x^cross(2)(x^2+1))/((x+y-1)(cross(x+y+1))))}}}

All that's left is

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

You can leave it like that or multiply out, and get

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

Edwin</pre>