Question 152643
First factor {{{n^3-27}}}



{{{n^3-27}}} Start with the numerator.



{{{(n)^3-(3)^3}}} Rewrite {{{n^3}}} as {{{(n)^3}}}. Rewrite {{{27}}} as {{{(3)^3}}}.



{{{(n-3)((n)^2+(n)(3)+(3)^2)}}} Now factor by using the difference of cubes formula. Remember the <a href="http://www.purplemath.com/modules/specfact2.htm">difference of cubes formula</a> is {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}



{{{(n-3)(n^2+3n+9)}}} Multiply



So {{{n^3-27}}} factors to {{{(n-3)(n^2+3n+9)}}}.


In other words, {{{n^3-27=(n-3)(n^2+3n+9)}}}



So {{{(n^3-27)/(n-3)}}} transforms into {{{((n-3)(n^2+3n+9))/(n-3)}}}



{{{((n-3)(n^2+3n+9))/(n-3)}}} Start with the given expression.



{{{(highlight((n-3))(n^2+3n+9))/highlight((n-3))}}} Highlight the common terms. 



{{{(cross((n-3))(n^2+3n+9))/cross((n-3))}}} Cancel out the common terms. 



{{{n^2+3n+9}}} Simplify



So {{{(n^3-27)/(n-3)}}} simplifies to {{{n^2+3n+9}}}



In other words, {{{(n^3-27)/(n-3)=n^2+3n+9}}} where {{{x<>3}}}