# Cube Root times Cube Root

• Nov 30th 2009, 10:03 AM
sharkman
Cube Root times Cube Root
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)

I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.

Is this the correct way to write the original problem another way?

I finally got (432m^4n^2)^2/3.

Is this correct?
• Nov 30th 2009, 10:15 AM
craig
Quote:

Originally Posted by sharkman
cube root of 16(m^2)(n) multiplied by cube root of 27(m^2)(n)

I converted the problem to (16m^2n)^1/3 * (27m^2n)^/3.

Is this the correct way to write the original problem another way?

I finally got (432m^4n^2)^2/3.

Is this correct?

Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$

In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.

Using the fact I gave first, this means than the answer is...

Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.
• Nov 30th 2009, 10:26 AM
sharkman
Quote:

Originally Posted by craig
Nearly there. Recall that $\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$

In your question you have $\sqrt[3]{16(m^2)(n)} \times \sqrt[3]{27(m^2)(n)}$.

Using the fact I gave first, this means than the answer is...

Hint: Your multiplication of the $m$ and $n$ terms was not wrong, it was the index - $\frac{2}{3}$ - that you got wrong.

The answer should be cubert{432m^4n^2}, right?
• Nov 30th 2009, 10:27 AM
craig
Correct :)
• Nov 30th 2009, 10:28 AM
sharkman
great
Quote:

Originally Posted by craig
Correct :)

Great! Thanks!
• Dec 1st 2009, 04:04 AM
HallsofIvy
Quote:

Originally Posted by sharkman
The answer should be cubert{432m^4n^2}, right?

Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.

The best way to write your answer is $6m\sqrt[3]{2mn^2}$.
• Dec 1st 2009, 07:03 AM
sharkman
ok
Quote:

Originally Posted by HallsofIvy
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $m^4= m^3 m$ so $\sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.

The best way to write your answer is $6m\sqrt[3]{2mn^2}$.

I understand. You decided to break the problem down a bit more.