Distributing

**cmf0106** · August 21st 2008, 09:13 AM

Using the distributive law of multiplication over subtraction i have a question

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

Why is the answer not $18y\sqrt[3]x-6x\sqrt[3]{x}$ ?
Because in the first half of the answer $18\sqrt[3]xy$ the x is under the radicand. While in the second half $6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

Please Help Clarify & Many Thanks

**Jhevon** · August 21st 2008, 09:19 AM

Originally Posted by cmf0106

Using the distributive law of multiplication over subtraction i have a question

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

Why is the answer not $18y\sqrt[3]x-6x\sqrt[3]{x}$ ?
Because in the first half of the answer $18\sqrt[3]xy$ the x is under the radicand. While in the second half $6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

Please Help Clarify & Many Thanks

you have the same thing
$\sqrt[3]{x}y = y \sqrt[3]{x}$ since multiplication is commutative

you are not to simplify $x \sqrt[3]{x}$ ?

**Chris L T521** · August 21st 2008, 09:23 AM

Originally Posted by cmf0106

Using the distributive law of multiplication over subtraction i have a question

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

Why is the answer not $18y\sqrt[3]x-6x\sqrt[3]{x}$ ?

That is the answer. The order of the variables in each term shouldn't change the answer; however, when we have expressions where there are terms that contain x and y, we usually write x first, and then y.

$18y\sqrt[3]{x}$ is correct, but so is this $18\sqrt[3]xy$

Because in the first half of the answer $18\sqrt[3]xy$ the x is under the radicand. While in the second half $6x\sqrt[3]{x}$ the x is outside the radicand. I do not understand why one x would be included under the radicand while the other would not.

Please Help Clarify & Many Thanks

Well, in the second term, the x term is inside and outside the radical.

$3\sqrt[3]{x}(6y-2x)\implies \left[3\sqrt[3]{x}\cdot 6y\right]-\left[3\sqrt[3]{x}\cdot 2x\right]\implies 18\sqrt[3]xy-6x\sqrt[3]x$

We could, however, combine $x$ and $\sqrt[3]x$ , but I believe that is beyond what the question is asking.

Does this help?

--Chris

**cmf0106** · August 21st 2008, 09:36 AM

$x\sqrt[3]{x}$ Not it does not ask me to, but im curious as to how to simply this. Im guessing its Using $\sqrt[n]{a^m}=a^{\frac{m}{n}}$ . But im not sure how it works out with another x right beside it.

EDIT: And yes it does help chris, but i would like to go beyond what the question is asking to enhance my knowledge.

**Chris L T521** · August 21st 2008, 09:38 AM

Originally Posted by cmf0106

$x\sqrt[3]{x}$ Not it does not ask me to, but im curious as to how to simply this. Im guessing its Using $\sqrt[n]{a^m}=a^{\frac{m}{n}}$ . But im not sure how it works out with another x right beside it.