How do you factor 3-8u^3?
\(a^3-b^3=(a-b)(a^2+ab+b^2)\\ \\ a=3^{1/3}\qquad b=2u\\ \\ 3-8u^{1/3}\;=\;(3^{1/3}-2u)(3^{2/3}+3^{1/3}*2u+4u^2)\)
