sqrt[4]{6} times sqrt[4]{36} times sqrt[4]{192}

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sqrt[4]{6} times sqrt[4]{36} times sqrt[4]{192}

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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lol, more of a fail -.- So, this is the question: \[\sqrt[4]{6} \times \sqrt[4]{36} \times \sqrt[4]{192}\]
I know it looks easy, I just keep on getting different answer :/
O.o

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I got \[2\sqrt[4]{2592}\] Yeah -.- It looks wrong -.-
If it has the same root....
rewrite it like this: \[\large \sqrt[4]{6 \times 36 \times 192}\]
Yes, this is what I did..
  • 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
|dw:1347737858173:dw|
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
sorry. i got \[ \large 12\sqrt[4]{2} \]
  • phi
@Hero that was to @aroub not you. I know you know how to do it!
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.
Thank you phi and everyone!! =D
  • 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)
If i still want the square root that would be equal to \[12\sqrt[4]{2}?\]
  • phi
yes, but it is the 4th root (not the square root)
Oh yeah sorry -.- I'm used to square roots :P

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