# Distributing

• Aug 21st 2008, 10:13 AM
cmf0106
Distributing
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.

• Aug 21st 2008, 10:19 AM
Jhevon
Quote:

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.

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

you are not to simplify $x \sqrt[3]{x}$ ?
• Aug 21st 2008, 10:23 AM
Chris L T521
Quote:

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 $18\sqrt[3]xy$

Quote:

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.

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
• Aug 21st 2008, 10:36 AM
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.

EDIT: And yes it does help chris, but i would like to go beyond what the question is asking to enhance my knowledge.
• Aug 21st 2008, 10:38 AM
Chris L T521
Quote:

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.

Keep in mind $x^a\cdot x^b=x^{a+b}$ where we have $x=x^1$ and $\sqrt[3]x=x^{\frac{1}{3}}$

So, $x\cdot x^{\frac{1}{3}}=\dots$

--Chris
• Aug 21st 2008, 11:04 AM
cmf0106
so $x^{1\frac{1}{3}}$?
• Aug 21st 2008, 11:06 AM
Chris L T521
Quote:

Originally Posted by cmf0106
so $x^{1\frac{1}{3}}$?

yes...or as an improper fraction $x^{\frac{4}{3}}$. In radical notation it would be $\sqrt[3]{x^4}$

Does this make more sense now?

--Chris
• Aug 21st 2008, 11:30 AM
cmf0106
Yes it does thanks for answering my questions Chris and everyone else!