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 $A+B+C$.
Since the cube of $5$ is $125$, we multiply the top and bottom by $\sqrt[3]{\frac{125}{5}}$, which is $\sqrt[3]{25}$. $$\frac{3}{2\sqrt[3]{5}} \cdot \frac{\sqrt[3]{25}}{\sqrt[3]{25}}=$$$$\frac{3\sqrt[3]{25}}{10}$$Therefore, $A+B+C=3+25+10=\boxed{38}$.
38
Level 4
Algebra
Problem: 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 $A+B+C$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To rationalize the denominator of the expression $\frac{3}{2\sqrt[3]{5}}$, we need to eliminate the cube root in the denominator. The denominator is $2\sqrt[3]{5}$, which can be expressed as $2 \cdot 5^{1/3}$. To rationalize it, we multiply both the numerator and the denominator by $\sqrt[3]{25}$, which is $5^{2/3}$. This choice is made because multiplying $5^{1/3}$ by $5^{2/3}$ results in $5^{(1/3 + 2/3)} = 5^{1} = 5$, a rational number.

The multiplication yields:

Numerator: 
$3 \cdot \sqrt[3]{25} = 3\sqrt[3]{25}$.

Denominator: 
$2\sqrt[3]{5} \cdot \sqrt[3]{25} = 2 \cdot 5^{1/3} \cdot 5^{2/3} = 2 \cdot 5 = 10$.

Thus, the expression simplifies to:

$\frac{3\sqrt[3]{25}}{10}$.

In this form, we identify $A = 3$, $B = 25$, and $C = 10$. 

Next, we check the conditions: $C$ is positive, and $B = 25$ is not divisible by the cube of any prime since $25 = 5^2$ and does not contain $5^3$. 

Finally, we compute $A + B + C$:

$A + B + C = 3 + 25 + 10 = 38$.

The final answer is $38$.<|END_OF_LATENT|> The final answer is $38$.