Question 74451
Factor completely:
{{{8x^3-27}}} Do you recognise this as the difference of two cubes?
{{{(2x)^3-(3)^3}}} There is a form for factoring the difference of two cubes:
{{{A^3-B^3 = (A-B)(A^2+AB+B^2)}}} Applying this to your case: {{{A=2x}}} and {{{B = 3}}} 
{{{(2x)^3-(3)^3 = (2x-3)((2x)^2+(2x)(3)+3^2)}}} Simplifying this:
{{{8x^3-27 = (2x-3)(4x^2+6x+9)}}} This is it!
Sometimes, it is possible to factor the second set of parentheses but in this case, that's not possible.