Rationalize the denominator: $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}$$
First, simplify $\sqrt[3]{16}=\sqrt[3]{2^3\cdot2}=2\sqrt[3]{2}$.  Substituting this, the fraction becomes: $$\frac{1}{\sqrt[3]{2}+2\sqrt[3]{2}}=\frac{1}{3\sqrt[3]{2}}$$ To rationalize this, we need to multiply the numerator and denominator by something that will eliminate the cube root in the denominator.  If we multiply $\sqrt[3]{2}$, by $\sqrt[3]{4}$, then the result will be $\sqrt[3]{2}\cdot\sqrt[3]{4}=\sqrt[3]{2\cdot4}=\sqrt[3]{8}=2$.  So, multiply the expression above by $\dfrac{\sqrt[3]{4}}{\sqrt[3]{4}}$. $$\frac{1}{3\sqrt[3]{2}}\cdot\frac{\sqrt[3]{4}}{\sqrt[3]{4}}=\frac{\sqrt[3]{4}}{3\sqrt[3]{8}}=\boxed{\frac{\sqrt[3]{4}}{6}}$$
\frac{\sqrt{[}3]{4}}{6}
Level 5
Algebra
Problem: Rationalize the denominator: $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To rationalize the denominator, first simplify the expression in the denominator. The denominator is $\sqrt[3]{2} + \sqrt[3]{16}$. Recognize that $\sqrt[3]{16}$ can be rewritten as $\sqrt[3]{2^4} = 2 \cdot \sqrt[3]{2}$. Thus, the denominator becomes $\sqrt[3]{2} + 2\sqrt[3]{2} = 3\sqrt[3]{2}$.

Next, rewrite the original expression using this simplification: 
$$\frac{1}{\sqrt[3]{2} + \sqrt[3]{16}} = \frac{1}{3\sqrt[3]{2}}.$$

To rationalize the denominator, multiply both the numerator and the denominator by $\sqrt[3]{4^2} = \sqrt[3]{16}$, which is the cube of the denominator's base. This gives:
$$\frac{1 \cdot \sqrt[3]{16}}{3\sqrt[3]{2} \cdot \sqrt[3]{16}} = \frac{\sqrt[3]{16}}{3\sqrt[3]{32}}.$$

Now, simplify the denominator. Since $\sqrt[3]{32} = \sqrt[3]{2^5} = 2\sqrt[3]{4}$, the expression becomes:
$$\frac{\sqrt[3]{16}}{3 \cdot 2\sqrt[3]{4}} = \frac{\sqrt[3]{16}}{6\sqrt[3]{4}}.$$

Next, simplify $\sqrt[3]{16}$. Since $16 = 4^2$, we have $\sqrt[3]{16} = \sqrt[3]{(4^2)} = \sqrt[3]{4}^2$. Thus, the expression can be rewritten as:
$$\frac{\sqrt[3]{4}^2}{6\sqrt[3]{4}}.$$

Now, cancel one $\sqrt[3]{4}$ from the numerator and the denominator:
$$\frac{\sqrt[3]{4}}{6}.$$

This leads to the final answer:
$$\frac{\sqrt[3]{4}}{6}.$$<|END_OF_LATENT|> The final answer is $\frac{\sqrt{[}3]{4}}{6}$.