anonymous
  • anonymous
how do i solve the cubed root of 3xcubed ytothe 5?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
you can write that as cuberoot(xxxyyyyy); then take out the triples
anonymous
  • anonymous
what about the cubed root of 24x^15 y^6?
anonymous
  • anonymous
so take out 3 of the x and 3 of the y and that leaves two y, so \[xy \sqrt[3]{y^2}\]

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anonymous
  • anonymous
ok i had an answer similar to that, but not that exact one.
anonymous
  • anonymous
take out as many triples as you can; and factor 24 into prime factors
anonymous
  • anonymous
im not understanding the triples thing.
anonymous
  • anonymous
24 is 2x2x2x3
anonymous
  • anonymous
exactly... there are 3 2s so they can come out
anonymous
  • anonymous
cuberoot(24) is 2 times cuberoot(3)
anonymous
  • anonymous
fifteen xs so you can pull 5 triples out (all of them)
anonymous
  • anonymous
so its 2x^5
anonymous
  • anonymous
on the outside?
anonymous
  • anonymous
or is the answer 2x^5 y^2?
anonymous
  • anonymous
so far, but dont forget the ys
anonymous
  • anonymous
yes... but there is something left under the radical
anonymous
  • anonymous
3?
anonymous
  • anonymous
yes
anonymous
  • anonymous
you got it.... for square roots you pull out pairs, for cube roots you pull out the triples
anonymous
  • anonymous
so the answer is 2x^5 y^2 cube root 3?
anonymous
  • anonymous
I really appreciate your help. There is one more I need help with. It is under a square root, but is a fraction
anonymous
  • anonymous
but be careful... this is the way to do single terms like these, but \[\sqrt{x^6 + y^2}\] is not x^3+y
anonymous
  • anonymous
yes, that is the answer... dnag you are good and let's see the fraction
anonymous
  • anonymous
thank you!
anonymous
  • anonymous
I'm not sure how to write it, but its cube root of 72x^15 y^13/64x^12 y^9
anonymous
  • anonymous
\[\sqrt[3]{72x^15y^13 \over 64x^12y^9}\]
anonymous
  • anonymous
simplify it first ... write the numbers as prime factors and cancel as much as you can
anonymous
  • anonymous
i need some help with that. I don't understand.
anonymous
  • anonymous
you can cancel 12 xs and 9 ys right off the bat
anonymous
  • anonymous
so is it cube root of 72x^3 y^4/64xy?
anonymous
  • anonymous
\[{x^{15} \over x^{12} } = x^3\]
anonymous
  • anonymous
15 - 12 leaves 3 xs, and they happened to be on top
anonymous
  • anonymous
so there are none left on the bottom, just the 64?
anonymous
  • anonymous
exactly... got lucky this time
anonymous
  • anonymous
with the problem I mean, you are doing great
anonymous
  • anonymous
thank you.
anonymous
  • anonymous
so then i factor 72 and 64 or?
anonymous
  • anonymous
now cancel out the common factors in 72 and 64
anonymous
  • anonymous
yes, thats how to do it
anonymous
  • anonymous
3x3x2x2x2?
anonymous
  • anonymous
thats 72 it seems
anonymous
  • anonymous
3x3x2x2x2 / 2x2x2x2x2x2 3x3 - - - - - - 2x2x2
anonymous
  • anonymous
2x2x3x3?
anonymous
  • anonymous
i know that last one is wrong.
anonymous
  • anonymous
keep dividing by 2 until you can't do it any more then see if what is left is prime
anonymous
  • anonymous
idk
anonymous
  • anonymous
64 is 2x2x2x2x2x2
anonymous
  • anonymous
right so that is 3
anonymous
  • anonymous
because there are 3 sets right?
anonymous
  • anonymous
not yet, we are trying to simplify... cancelling what we can
anonymous
  • anonymous
oh
anonymous
  • anonymous
that 72 is still up there, I mean the 3x3x2x2x2
anonymous
  • anonymous
so 3x3x2x2x2 / 2x2x2x2x2x2 is 3x3 - - - - - - 2x2x2
anonymous
  • anonymous
cacelled out 2x2x2 from top and bottom
anonymous
  • anonymous
ok
anonymous
  • anonymous
so that leaves 9 x^3y^4 over 8 in the radical
anonymous
  • anonymous
what goes on the outside?
anonymous
  • anonymous
we havent pulled anything out yet, just simplified it you pick the triples to pull out (since this is a cube root)
anonymous
  • anonymous
is it just one y on the outside?
anonymous
  • anonymous
\[\sqrt[3]{9x^3y^4 \over 8}\]
anonymous
  • anonymous
yes, pull 3ys out, one ouside and leaves one inside
anonymous
  • anonymous
\[3y^{3}\] and \[x^{3}\] on the outside
anonymous
  • anonymous
and then 9y/8 on the inside?
anonymous
  • anonymous
almost... but when you take 3 of something from inside, it is 1 on the outside so an x and a y and the 1/8 can come out
anonymous
  • anonymous
which does leave 9y inside ... and the cube root of 1/8 is 1/2
anonymous
  • anonymous
1/8xy on the outside
anonymous
  • anonymous
1/2 x y on the outside
anonymous
  • anonymous
x is the cube root of x cubed .... y is the cube root of y cubed .... 1/2 is the cube root of 1/8
anonymous
  • anonymous
so an x outside represents 3 inside, and so on
anonymous
  • anonymous
ok so then its just x^2 y^2 on the inside?
anonymous
  • anonymous
just a y on the inside?
anonymous
  • anonymous
I think the answer is: \[{1 \over 2}xy \sqrt[3]{9y}\]
anonymous
  • anonymous
you are right.
anonymous
  • anonymous
you can check it by cubing what is on the left and multiplying that by what is under the radical
anonymous
  • anonymous
there are lots of shortcuts, but they always confuse me I like doing things the long simple way
anonymous
  • anonymous
okay great. thank you so much for your help. that was a hard one.
anonymous
  • anonymous
only because of the typing... you knwo what you are doing keep up the good work
anonymous
  • anonymous
thank you.

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