Hi! Could someone explain how to solve this? \[\sqrt[3]{4}\times \sqrt[2]{8}\] thank you! <3

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Hi! Could someone explain how to solve this? \[\sqrt[3]{4}\times \sqrt[2]{8}\] thank you! <3

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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Hm. Is that right or is it the opposite? (\(\sqrt[3]{8} \times \sqrt[2]{4}\) ) ?
sorry, for some reason, the equation writer won't work. and nope, it's not. :D the answer according to my answer key is 4 sqrt[6]{2), but i don't know how to get it.
If the problem is typed correctly, use index notation and put in powers of 2: cubrt4 = 4^(1/3) = ((2^2)^(1/3) = 2^(2/3) sqrt8 = ((2^3)^(1/2)) = 2^(3/2) Multiply the two by adding indices to get 2^(7/6) which is not the same as your answer, unfortunately

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i'm really sure that the answer is \[4\sqrt[6]{2}\] i even checked it with my calculator myself. but i just don't know how to get it. :(
\[\sqrt[3]{4}\sqrt{8} \]Square the expression\[\left(\sqrt[3]{4}\sqrt{8}\right)^2=16\ 2^{1/3} \]Take the square root of the result.\[\sqrt{16\ 2^{1/3}}=4\ 2^{1/6}=4 \sqrt[6]{2} \]
Hero in da heezy. Nice one robtobey.
Aw shucks. Thanks.

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