Question 328292
27-8u^3
(3)^3-(2u)^3
This is difference of cubes.
the general formula is {{{a^3-b^3=(a-b)(a^2+ab+b^2)}}} 
a=3 & b= 2u
So our solution will be {{{(3-2u)(3^2+3*2u+(2u)^2)}}}
{{{(3-2u)(9+6u+4u^2)}}}