Simplify $\sqrt[3]{1+8} \cdot \sqrt[3]{1+\sqrt[3]{8}}$.
The first cube root becomes $\sqrt[3]{9}$. $\sqrt[3]{8}=2$, so the second cube root becomes $\sqrt[3]{3}$. Multiplying these gives $\sqrt[3]{27} = \boxed{3}$.
3
Level 2
Algebra
Problem: Simplify $\sqrt[3]{1+8} \cdot \sqrt[3]{1+\sqrt[3]{8}}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first evaluate the individual cube roots. Start with $\sqrt[3]{1+8}$, which simplifies to $\sqrt[3]{9}$. Next, consider $\sqrt[3]{1+\sqrt[3]{8}}$. Since $\sqrt[3]{8} = 2$, this becomes $\sqrt[3]{1+2} = \sqrt[3]{3}$. 

Now, the expression can be rewritten as $\sqrt[3]{9} \cdot \sqrt[3]{3}$. Using the property of cube roots that states $\sqrt[3]{a} \cdot \sqrt[3]{b} = \sqrt[3]{a \cdot b}$, we combine the two cube roots: 

$\sqrt[3]{9} \cdot \sqrt[3]{3} = \sqrt[3]{9 \cdot 3} = \sqrt[3]{27}$. 

Next, simplify $\sqrt[3]{27}$. Since $27$ is $3^3$, we find that $\sqrt[3]{27} = 3$. 

Thus, the final answer is $3$.<|END_OF_LATENT|> The final answer is $3$.