Question

Simplify the radical. 16xy6−−−−−√316xy63 (1 point) Responses 2xy22–√32xy2232 x y squared cube root of 2 2y22x−−√32y22x32 y squared cube root of 2 x 8y22x−−√38y22x38 y squared cube root of 2 x 4y2x−−√3

Answer 1

To simplify the expression \(\sqrt{16xy^6} \div \sqrt{316xy^3}\), we first simplify each term separately.

### Step 1: Simplify \(\sqrt{16xy^6}\)
\[
\sqrt{16xy^6} = \sqrt{16} \cdot \sqrt{x} \cdot \sqrt{y^6}
\]
Calculating each part:
- \(\sqrt{16} = 4\)
- \(\sqrt{y^6} = y^{6/2} = y^3\)

Thus,
\[
\sqrt{16xy^6} = 4y^3\sqrt{x}
\]

### Step 2: Simplify \(\sqrt{316xy^3}\)
\[
\sqrt{316xy^3} = \sqrt{316} \cdot \sqrt{x} \cdot \sqrt{y^3}
\]
Note that \(316 = 4 \cdot 79\):
- \(\sqrt{316} = \sqrt{4 \cdot 79} = \sqrt{4} \cdot \sqrt{79} = 2\sqrt{79}\)
- \(\sqrt{y^3} = y^{3/2} = y^{1} \cdot \sqrt{y} = y\sqrt{y}\)

Thus,
\[
\sqrt{316xy^3} = 2\sqrt{79} \cdot \sqrt{x} \cdot y\sqrt{y} = 2y\sqrt{79xy}
\]

### Step 3: Combine the results
Now we combine this into the original expression:
\[
\frac{\sqrt{16xy^6}}{\sqrt{316xy^3}} = \frac{4y^3\sqrt{x}}{2y\sqrt{79xy}} = \frac{4y^3}{2y} \cdot \frac{\sqrt{x}}{\sqrt{79xy}} = 2y^2 \cdot \frac{\sqrt{x}}{\sqrt{79}\sqrt{xy}}
\]

### Step 4: Simplify further
This reduces to:
\[
2y^2 \cdot \frac{\sqrt{x}}{\sqrt{79}\sqrt{xy}} = 2y^2 \cdot \frac{\sqrt{x}}{\sqrt{79}\sqrt{x}\sqrt{y}} = 2y^2 \cdot \frac{1}{\sqrt{79 \cdot y}} = \frac{2y^2}{\sqrt{79y}}
\]

### Final result
Thus, the simplified form is:
\[
\frac{2y^2}{\sqrt{79y}}
\]

This result rationalizes and simplifies the original radical expression.