Question

Simplify the radical.
3
V 16xy6
(1 point)
○ 2xy2√2
○ 8y2V/2x
○ 2y2/2x
○ 4y² Vx
4y²/x

Answer 1

To simplify the radical \( \sqrt[3]{16xy^6} \), we can start by breaking it down into its prime factors.

1. Factor out the numbers and variables:
- \( 16 = 2^4 \)
- \( x \) remains \( x \)
- \( y^6 \) can be written as \( (y^2)^3 \)

So we rewrite the expression:
\[
\sqrt[3]{16xy^6} = \sqrt[3]{2^4 \cdot x \cdot (y^2)^3}
\]

2. Split the cube root:
\[
\sqrt[3]{2^4} \cdot \sqrt[3]{x} \cdot \sqrt[3]{(y^2)^3}
\]
\[
= \sqrt[3]{2^4} \cdot \sqrt[3]{x} \cdot y^2
\]

3. Simplify \( \sqrt[3]{2^4} \):
- \( 2^4 = 2^3 \cdot 2 = 8 \cdot 2 \)
- Therefore, \( \sqrt[3]{2^4} = \sqrt[3]{2^3 \cdot 2} = \sqrt[3]{2^3} \cdot \sqrt[3]{2} = 2 \cdot \sqrt[3]{2} \)

So now we have:
\[
2 \cdot \sqrt[3]{2} \cdot y^2 \cdot \sqrt[3]{x}
\]

Combining these gives:
\[
2y^2 \sqrt[3]{2x}
\]

Now we check the options provided:
- \( 2xy^2 \sqrt{2} \)
- \( 8y^2 \sqrt{2/x} \)
- \( \frac{2y^2}{2x} \)
- \( 4y^2 \sqrt{x} \)
- \( \frac{4y^2}{x} \)

None of the provided options exactly match the simplified form \( 2y^2 \sqrt[3]{2x} \). Please double-check the options or the original expression for a possible error or a different configuration!