Question 146834
Start with {{{(81-x^4)/(3-x)}}} divided by {{{(9+x^2)/(9)}}}.  


Find the common denominator, which would be 9(3-x).  Manipulate the fractions and get {{{(9(81-x^4))/9(3-x)}}} divided by {{{((3-x)(9+x^2))/(9(3-x))}}}.  


Then eliminate the denominators from the equation.  Get {{{9(81-x^4)}}} divided by {{{(3-x)(9+x^2)}}}.  


Factor all of the expressions.  The expression {{{81-x^4}}} factors into {{{(9+x^2)(9-x^2)}}}.  Then {{{9-x^2}}} factors to {{{(3-x)(3+x)}}}.  So the numerator becomes {{{9(9+x^2)(3-x)(3+x)}}} while the denominator remains {{{(9+x^2)(3-x)}}}. So it looks like: {{{(9(9+x^2)(3-x)(3+x))/((9+x^2)(3-x))}}}. 


Now just cancel out like terms in the numerator and the denominator, which would be {{{(3-x)}}} and {{{(9+x^2)}}}. Then you are left with {{{(9(3+x))/1}}}.  or {{{9(3+x)}}}.  You can stop there if you want to. Going further is not necessary, but if you want to anyway, read on.


It is very easy from here.  Just distribute the 9 to both terms and get {{{(27+9x)/1}}} or simply {{{27+9x}}}.