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 $A+B+C$.
To rationalize the denominator, I need to multiply the fraction by something that will make the cube root go away. I can use the fact that $(\sqrt[3]{5})^3=5$ to help me find that something. If I multiply the fraction by $\frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}$, which is just 1, I will get $\frac{3\sqrt[3]{5^2}}{2\sqrt[3]{5}\sqrt[3]{5^2}}=\frac{3\sqrt[3]{5^2}}{2\sqrt[3]{5^3}}=\frac{3\sqrt[3]{5^2}}{2\cdot 5}$. Now I have rationalized the denominator, and the answer is in the form of $\frac{A\sqrt[3]{B}}{C}$, where $A=3$, $B=5^2$, and $C=2\cdot 5=10$. Therefore, $A+B+C=3+25+10=38$. 