Question

HELP ME AND DO NOT GUESS PLEASE HELP ME I AM DYING'

Answer 1

$\sqrt[3]{ \frac{12x^2}{16y} } = \sqrt[3]{ \frac{3x^2}{4y} } = \frac{ \sqrt[3]{3x^2} }{ \sqrt[3]{4y} }= \frac{ \sqrt[3]{3x^2} \cdot \sqrt[3]{(4y^2)} }{ \sqrt[3]{4y} \cdot \sqrt[3]{(4y^2)} }=\frac{ \sqrt[3]{48x^2y^2} }{ \sqrt[3]{64y^3} }= \frac{ 2\sqrt[3]{6x^2y^2} }{ 4y}= \\\\\\ = \frac{ \sqrt[3]{6x^2y^2} }{ 2y}$

Answer 2

Answer:

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

Step-by-step explanation:

The given expression is:

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

Upon solving the above expression, we have

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

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

Now, multiplying and dividing by $\sqrt[3]{(4y)^2}$, we have

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

=$\frac{\sqrt[3]{48x^2y^2}}{\sqrt[3]{64y^3}}$

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

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

which is the required simplified form of the above given expression.

Thus, option (E) is correct.