# Math Help - Basic Factoring Roots question

1. ## Basic Factoring Roots question

In the following expression

$3\sqrt[3]x(6y-2x)= 3\sqrt[3]x(6y)-3\sqrt[3]x(2x)=
18\sqrt[3]{xy}-6x\sqrt[3]x$

I do not understand how to decide which number ends up under the radical sign.

For instance in the first term $3\sqrt[3]x(6y)=
18\sqrt[3]{xy}-6x\sqrt[3]x$

Where as in the second term $3\sqrt[3]x(2x)=6x\sqrt[3]x$ the x goes outside the radical instead of under like the above part of the expression. The author does not explain this process and its leaving me very confused.

2. Originally Posted by allyourbass2212
In the following expression

$3\sqrt[3]x(6y-2x)= 3\sqrt[3]x(6y)-3\sqrt[3]x(2x)=
18\sqrt[3]{xy}-6x\sqrt[3]x$

I do not understand how to decide which number ends up under the radical sign.

For instance in the first term $3\sqrt[3]x(6y)=
18\sqrt[3]{xy}-6x\sqrt[3]x$
It shouldn't. That's a mistake. The y should remain outside the radical:
$3\sqrt[3]x(6y)=18y\sqrt[3]{x}$

If the y was under the radical to begin with, like this:
$3\sqrt[3]x(6{\color{red}\sqrt[3]{y}}-2x)= 3\sqrt[3]x(6{\color{red}\sqrt[3]{y}})-3\sqrt[3]x(2x)=
18\sqrt[3]{xy}-6x\sqrt[3]x$

then the y should remain under the radical:
$3\sqrt[3]x(6\sqrt[3]{y})=18\sqrt[3]{xy}$

Where as in the second term $3\sqrt[3]x(2x)=6x\sqrt[3]x$ the x goes outside the radical instead of under like the above part of the expression.
This is correct.

01

3. This books is horrible, full of errors and very brief on explanations.

At any rate should the variable always go outside the radical in such cases?

$3\sqrt[3]x(6{\color{red}y})=18{\color{red}y}\sqrt[3]{x}$