Question

1. simplify. $sqrt3{4xy^2} cdot sqrt3{16x^2y^4} cdot sqrt3{2x^4y^6}$

$4xy^4sqrt3{2x^2}$
$2x^2y^4sqrt3{2x}$
$4x^2y^4sqrt3{2x}$
$2x^2y^2sqrt3{2xy^2}$

Explanation:

Step1: Combine cube roots

$\sqrt[3]{4xy^2 \cdot 16x^2y^4 \cdot 2x^4y^6}$

Step2: Multiply coefficients inside root

$\sqrt[3]{(4 \cdot 16 \cdot 2)x^{1+2+4}y^{2+4+6}} = \sqrt[3]{128x^7y^{12}}$

Step3: Factor into perfect cubes

$\sqrt[3]{64 \cdot 2 \cdot x^6 \cdot x \cdot y^{12}} = \sqrt[3]{64x^6y^{12} \cdot 2x}$

Step4: Extract perfect cubes

$\sqrt[3]{64x^6y^{12}} \cdot \sqrt[3]{2x} = 4x^2y^4\sqrt[3]{2x}$

Answer:

$4x^2y^4\sqrt[3]{2x}$