# Thread: Cube Root times Cube Root

1. ## 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?

2. 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 $\displaystyle \sqrt{a} \times \sqrt{b} = \sqrt{ab}$

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

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

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

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.
The answer should be cubert{432m^4n^2}, right?

4. Correct 5. ## great Originally Posted by craig Correct Great! Thanks!

6. 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. $\displaystyle m^4= m^3 m$ so $\displaystyle \sqrt{m^4}= \sqrt{m^3}\sqrt{m}= m\sqrt{m}$. Perhaps more importantly, $\displaystyle 432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\displaystyle \sqrt{432}= \sqrt{2^3}\sqrt{3^3}\sqrt(2)= 6\sqrt{2}$.

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

7. ## ok 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. $\displaystyle m^4= m^3 m$ so $\displaystyle \sqrt{m^4}= \sqrt{m^3}\sqrt{m}= m\sqrt{m}$. Perhaps more importantly, $\displaystyle 432= (16)(27)= (2^4)(3^3)= (2^3)(3^3)(2)$ so $\displaystyle \sqrt{432}= \sqrt{2^3}\sqrt{3^3}\sqrt(2)= 6\sqrt{2}$.

The best way to write your answer is $\displaystyle 6m\sqrt{2mn^2}$.
I understand. You decided to break the problem down a bit more.

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