Question 607498
Sum of cubes:
{{{a^3 + b^3 = (a+b)(a^2-ab+b^2)}}}
.
Factoring numerator:
{{{3x^3+24}}}
{{{3(x^3+8)}}}
{{{3(x^3+2^3)}}}
{{{3(x+2)(x^2-2x+2^2)}}}
{{{3(x+2)(x^2-2x+4)}}}

Factoring denominator:
{{{12-3x^2}}}
{{{3(4-x^2)}}}
{{{3(2^2-x^2)}}}
{{{3(2-x)(2+x)}}}
. 
So, now we can rewrite:
{{{(3x^3+24)/(12-3x^2)}}}
as
{{{(3(x+2)(x^2-2x+4))/(3(2-x)(2+x))}}}
{{{((x+2)(x^2-2x+4))/((2-x)(2+x))}}}
{{{(x^2-2x+4)/(2-x)}}}