Simplify, help please??!! This one's harder...
\[\LARGE 5\sqrt[3]{96}\]
@hartnn
Thanks, @Algebraic! He went offline because of you!!! He actually helped me & explained it! I am VERY upset with you -____-
it's all very sad.
you need to factor 96 and see if it contains any perfect cubes
;_; b
but thumbs up anyway.
@satellite73 I don't know how to do that... Will you please explain?
how to factor? sure
Btw, answer choices (sorry forgot): \[\large A) 20\sqrt[3]{6}\] \[\large B) 10\sqrt[3]{12}\] \[\large C) 7\sqrt[3]{12}\] \[\large D) 2\sqrt[3]{12}\]
but before we do that, lets see what the answer is, there are not too many perfect cubes to check \(2^3=8\) and \(3^3=27\) so lets check for 8 if your multiplication tables are very good then you may recognize \(96=8\times 12\) so \[\sqrt[3]{96}=\sqrt[3]{8}\sqrt[3]{12}=2\sqrt[3]{12}\]
meaning your "final answer" will be \[5\times 2\sqrt[3]{12}=10\sqrt[3]{12}\]
Wait, you confused me on that last part. Can you elaborate a little more, please?
idea is you are looking to see if 96 contains any perfect cubes do you know what i mean by "perfect cube"?
Yes
perfect cubes means the cube of some whole number let me list some \(2^3=8,3^3=27,4^3=64,5^3=125\) and so on
so now i am looking at \(\sqrt[3]{96}\) ignoring the 5 out front, because it is not important in this computation i want to know if there are any perfect cubes that are factors of 96 so i start with 8 (because it is the smallest)
i see that 8 goes in to 96 evenly, because \(96=8\times 12\)
and i know that the cube root of 8 is 2 (because i know \(2^3=8\) so i pull a 2 outside of the radical, and leave the 12 inside
Yes.
that tells me that \[\sqrt[3]{96}=2\sqrt[3]{12}\]
now i remember that i started with \[5\sqrt[3]{96}\] so i end up with \[5\times 2\sqrt[3]{12}=10\sqrt[3]{12}\]
Oh, okay! I think I understand now :) Thank you so much for taking the time to explain it to me. Sometimes it just doesn't process :)
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