## aroub 2 years ago sqrt[4]{6} times sqrt[4]{36} times sqrt[4]{192}

1. aroub

lol, more of a fail -.- So, this is the question: $\sqrt[4]{6} \times \sqrt[4]{36} \times \sqrt[4]{192}$

2. aroub

I know it looks easy, I just keep on getting different answer :/

3. aroub

O.o

4. aroub

I got $2\sqrt[4]{2592}$ Yeah -.- It looks wrong -.-

5. Hero

If it has the same root....

6. Hero

rewrite it like this: $\large \sqrt[4]{6 \times 36 \times 192}$

7. aroub

Yes, this is what I did..

8. phi

I would never multiply these numbers together. on the contrary, you want to factor them all are under the root sign. so you have $6 \cdot 6^2 \cdot 6\cdot 2^5$ or $6^4 \cdot 2^4 \cdot 2$ pull out the powers of 4

9. RaphaelFilgueiras

|dw:1347737858173:dw|

10. Hero

I didn't multiply them together. I just simplified them under one root @phi. I agree that what you did would be the next step AFTER putting them under one root

11. helder_edwin

sorry. i got $\large 12\sqrt[4]{2}$

12. phi

@Hero that was to @aroub not you. I know you know how to do it!

13. aroub

I didn't multiply them together at first. This is was what I did (it's stupid though compared to what you did): $\sqrt[4]{3\times2 \times 6^2\times2^6\times3} = \sqrt[4]{3^2\times2^4\times6^2\times2^3}$ Don't ask me how or why :P And then after realizing I cant factor them more I multiplied them.

14. aroub

Thank you phi and everyone!! =D

15. phi

doing that is fine. The only mistake is not factor the 6^2 into 2^2 * 3^2 after combining bases, you would have gotten 3^4 2^9 or 3^4 * 2^4 * 2^4 * 2 take the 4th root, you would get 3*2*2 * 2^(1/4) or 12* 2^(1/4)

16. aroub

If i still want the square root that would be equal to $12\sqrt[4]{2}?$

17. phi

yes, but it is the 4th root (not the square root)

18. aroub

Oh yeah sorry -.- I'm used to square roots :P