Question 1122918
Factor the sum 
{{{250r^3-128t^3}}}
The special factoring formula for the difference of cubes, namely
{{{(x^3-y^3)=(x-y)(x^2+xy+y^2)}}}
:
factor out 2, then use the dif of cubes formula
{{{(250r^3-128t^3)}}} = {{{2(125r^3-64t^3)}}} = {{{2(5r-4t)(25r^2 + 20rt + 16t^2)}}}