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?

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- Nov 30th 2009, 09:03 AMsharkmanCube 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, 09:15 AMcraig
Nearly there. Recall that $\displaystyle \sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{ab}$

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

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

Hint: Your multiplication of the $\displaystyle m$ and $\displaystyle n$ terms was not wrong, it was the index - $\displaystyle \frac{2}{3}$ - that you got wrong. - Nov 30th 2009, 09:26 AMsharkmananswer...
- Nov 30th 2009, 09:27 AMcraig
Correct :)

- Nov 30th 2009, 09:28 AMsharkmangreat
- Dec 1st 2009, 03:04 AMHallsofIvy
Don't be too shocked if your teacher marks it wrong because it has not been simplified as much as it could be. $\displaystyle m^4= m^3 m$ so $\displaystyle \sqrt[3]{m^4}= \sqrt[3]{m^3}\sqrt[3]{m}= m\sqrt[3]{m}$. Perhaps more importantly, $\displaystyle 432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\displaystyle \sqrt[3]{432}= \sqrt[3]{2^3}\sqrt[3]{3^3}\sqrt[3](2)= 6\sqrt[3]{2}$.

The best way to write your answer is $\displaystyle 6m\sqrt[3]{2mn^2}$. - Dec 1st 2009, 06:03 AMsharkmanok