Question

Simplify the radical. 16xy6−−−−−√3 (1 point) Responses 2y22x−−√3 2 y squared cube root of 2 x 2xy22–√3 2 x y squared cube root of 2 8y22x−−√3 8 y squared cube root of 2 x 4y2x−−√3

Answer 1

To simplify the expression \(\sqrt[3]{16xy^6}\), we can begin by breaking down the components inside the cube root.

1. **Factor the components**:
- \(16\) can be expressed as \(2^4\).
- \(x\) stays as \(x\).
- \(y^6\) is already a perfect cube since \(y^6 = (y^2)^3\).

So we rewrite the expression inside the cube root:
\[
16xy^6 = 2^4 x y^6 = 2^3 \cdot 2 \cdot x \cdot (y^2)^3
\]

2. **Apply the cube root**:
Now we can take the cube root:
\[
\sqrt[3]{16xy^6} = \sqrt[3]{2^3 \cdot 2 \cdot x \cdot (y^2)^3}
\]
This can be simplified as follows:
\[
= \sqrt[3]{2^3} \cdot \sqrt[3]{(y^2)^3} \cdot \sqrt[3]{2} \cdot \sqrt[3]{x}
\]
\[
= 2 \cdot y^2 \cdot \sqrt[3]{2} \cdot \sqrt[3]{x}
\]

3. **Combine under a single radical**:
We can express that as:
\[
= 2y^2 \sqrt[3]{2x}
\]

Thus, the simplified form of \(\sqrt[3]{16xy^6}\) is:
\[
\boxed{2y^2 \sqrt[3]{2x}}
\]

So the original expression \(16xy^6^{\frac{1}{3}} = 2y^2 \sqrt[3]{2x}\).