Question

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

What is the simplified form of the following expression? Assume

THE FIRST IMAGE IS THE ORIGINAL, THE LAST 4 ARE ANSWER

Answer 1

`\sqrt[3]{ (12x^2)/(16y) } = \sqrt[3]{ (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

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

Explanation:

The given expression is:

`\sqrt[3]{(12x^2)/(16y)}`

Upon solving the above expression, we have

=
`\sqrt[3]{(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}}{*}\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.