Question

What is the simplified form of the following expression? Assume y=0 ^3 sqrt 12x^2/16y

Answer 1

Answer: choice D

Step-by-step explanation: took it on edge

Answer 2

For this case we must simplify the following expression:

$\sqrt [3] {\frac {12x ^ 2} {16y}}$

We rewrite the expression as:

$\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\sqrt[3]{\frac{4(3x^2)}{4(4y)}}=\\\frac{\sqrt[3]{3x^2}}{\sqrt[3]{4y}}=$

We multiply the numerator and denominator by:

$(\sqrt[3]{4y})^2:\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{\sqrt[3]{4y}*(\sqrt[3]{4y})^2}=$

We use the rule of power$a ^ n * a ^ m = a ^ {n + m}$ in the denominator:

$\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{(\sqrt[3]{4y})^3}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{4y})^2}{4y}=$

Move the exponent within the radical:

$\frac{\sqrt[3]{3x^2}*(\sqrt[3]{16y^2}}{4y}=\\\frac{\sqrt[3]{3x^2}*(\sqrt[3]{2^3*(2y^2)}}{4y}=$

$\frac{2\sqrt[3]{3x^2}*(\sqrt[3]{(2y^2)}}{4y}=\\\frac{2\sqrt[3]{6x^2*y^2}}{4y}=$

$\frac{\sqrt[3]{6x^2*y^2}}{2y}$

Answer:

$\frac{\sqrt[3]{6x^2*y^2}}{2y}$