Question 174644
In this problem, you need to factor the numerator by applying the special factor condition known as the "sum of cubes".  For a complete explanation see:
http://www.purplemath.com/modules/specfact2.htm
.
{{{(8m^3 + 27)/(2m+3)}}}
is equivalent to:
{{{((2m)^3 + (3)^3)/(2m+3)}}}
.
Now, we can apply the "sum of cubes" rule of:
a^3 + b^3 = (a + b)(a^2 – ab + b^2) 
.
{{{(2m+3)((2m)^2-(2m)(3)+(3)^2)/(2m+3)}}}
The (2m+3) cancels leaving:
{{{(2m)^2-(2m)(3)+(3)^2}}}
Expanding:
{{{4m^2-6m+9}}}