Compute $\sqrt{54}\cdot\sqrt{32}\cdot \sqrt{6}$.
First, we simplify the radicals as much as possible. We have $\sqrt{54} = \sqrt{2\cdot 3^3} = \sqrt{2\cdot 3\cdot 3^2} = 3\sqrt{2\cdot 3} = 3\sqrt{6}$, and $\sqrt{32} = \sqrt{2^5} = \sqrt{2^4\cdot 2} = 4\sqrt{2}$.  Therefore, we have \begin{align*}\sqrt{54}\cdot\sqrt{32} \cdot \sqrt{6} &= (3\sqrt{6})(4\sqrt{2})(\sqrt{6}) = 3\cdot 4\sqrt{6}\cdot\sqrt{2}\sqrt{6}\\
&= 12\sqrt{2}(\sqrt{6}\sqrt{6}) = (12\sqrt{2})(6) = \boxed{72\sqrt{2}}.\end{align*}
72\sqrt{2}
Level 5
Prealgebra
Problem: Compute $\sqrt{54}\cdot\sqrt{32}\cdot \sqrt{6}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, first apply the property of square roots that states $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$. This allows us to combine the square roots into one expression. 

Calculate the product inside the square root: 
54 can be factored as $2 \cdot 27$ or $2 \cdot 3^3$. 
32 can be factored as $16 \cdot 2$ or $2^5$. 
6 can be factored as $2 \cdot 3$. 

Now, combine these factors:
$54 \cdot 32 \cdot 6 = (2 \cdot 3^3) \cdot (2^5) \cdot (2 \cdot 3) = 2^{1+5+1} \cdot 3^{3+1} = 2^7 \cdot 3^4$.

Next, take the square root of the combined product:
$\sqrt{2^7 \cdot 3^4} = \sqrt{2^7} \cdot \sqrt{3^4}$.

Calculate each square root:
$\sqrt{2^7} = \sqrt{2^6 \cdot 2} = \sqrt{2^6} \cdot \sqrt{2} = 2^3 \cdot \sqrt{2} = 8\sqrt{2}$.
$\sqrt{3^4} = 3^2 = 9$.

Now, multiply these results together:
$8\sqrt{2} \cdot 9 = 72\sqrt{2}$.

Thus, the final answer is $72\sqrt{2}$.<|END_OF_LATENT|> The final answer is $72\sqrt{2}$.