Distributing

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

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

Why is the answer not $\displaystyle 18y\sqrt[3]x-6x\sqrt[3]{x}$?
Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ 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 (Happy)

Quote:

---

Originally Posted by cmf0106

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

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

Why is the answer not $\displaystyle 18y\sqrt[3]x-6x\sqrt[3]{x}$?
Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ 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 (Happy)

---

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

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

Quote:

---

Originally Posted by cmf0106

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

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

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

---

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.

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

Quote:

---

Because in the first half of the answer $\displaystyle 18\sqrt[3]xy$ the x is under the radicand. While in the second half $\displaystyle 6x\sqrt[3]{x}$ 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 (Happy)

---

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

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

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

Does this help?

--Chris

$\displaystyle x\sqrt[3]{x}$ Not it does not ask me to, but im curious as to how to simply this. Im guessing its Using $\displaystyle \sqrt[n]{a^m}=a^{\frac{m}{n}}$. 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.

Quote:

---

Originally Posted by cmf0106

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

---

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

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

--Chris

so $\displaystyle x^{1\frac{1}{3}}$?

Quote:

---

Originally Posted by cmf0106

so $\displaystyle x^{1\frac{1}{3}}$?

---

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

Does this make more sense now?

--Chris

Yes it does thanks for answering my questions Chris and everyone else!