To determine which choice is equivalent to the product [tex]\(\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{8}\)[/tex], let's begin by understanding how to handle the multiplication of square roots.
First, recall the property of square roots that states:
[tex]\[
\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}
\][/tex]
Using this property, we can combine the square roots one step at a time:
[tex]\[
\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{8} = \sqrt{2 \cdot 3 \cdot 8}
\][/tex]
Now let's multiply the numbers under the square root:
[tex]\[
2 \cdot 3 = 6
\][/tex]
[tex]\[
6 \cdot 8 = 48
\][/tex]
So, we have:
[tex]\[
\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{8} = \sqrt{48}
\][/tex]
Next, we simplify [tex]\(\sqrt{48}\)[/tex]. To do this, we find the prime factorization of 48:
[tex]\[
48 = 2^4 \cdot 3
\][/tex]
We can break down [tex]\(\sqrt{48}\)[/tex] using the property that [tex]\(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\)[/tex]:
[tex]\[
\sqrt{48} = \sqrt{2^4 \cdot 3} = \sqrt{2^4} \cdot \sqrt{3}
\][/tex]
Since [tex]\(\sqrt{2^4} = 2^2 = 4\)[/tex], we get:
[tex]\[
\sqrt{48} = 4 \cdot \sqrt{3}
\][/tex]
Thus, the product [tex]\(\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{8}\)[/tex] simplifies to [tex]\(4 \sqrt{3}\)[/tex].
Therefore, the correct choice is:
[tex]\[
\boxed{4 \sqrt{3}}
\][/tex]
