Alg2: Cube root in the denominator?

Question

Let the &'s be radicals because i can't find how to type it in my computer.

^3&320& + ^3&1250&
-----------------------------------
2 ^3&5&

How can i get rid of the radical on the bottom?

Answer 1

((320)^(1/3) + (1250)^(1/3)) / 2(5)^(1/3) =
(4(5)^(1/3) + 5(10)^(1/3)) / 2(5)^(1/3) =
(4 + 5(2)^(1/3)) / 2 =
2 + (5/2)(2)^(1/3)