Rationalize the denominator of $\frac{3}{2\sqrt[3]{5}}$. The answer can be written in the form of $\frac{A\sqrt[3]{B}}{C}$, where $A$, $B$, and $C$ are integers, $C$ is positive, and $B$ is not divisible by the cube of any prime. Find $A+B+C$.
\($\frac{3}{2\sqrt[3]{5}}$ \) =
3 [ 5^(2/3) ]
_________ ________ =
2 * (5)^(1/3) [5^(2/3) ]
3 [ 5^(2/3) ] 3 *(5^2)^(1/3) 3 ∛25
___________ = ____________ = ______
2 * 5 10 10
A = 3 B = 25 C = 10
And their sum = 38
+21 
