how do i solve the cubed root of 3xcubed ytothe 5?

- anonymous

how do i solve the cubed root of 3xcubed ytothe 5?

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- anonymous

you can write that as cuberoot(xxxyyyyy); then take out the triples

- anonymous

what about the cubed root of 24x^15 y^6?

- 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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## More answers

- anonymous

ok i had an answer similar to that, but not that exact one.

- anonymous

take out as many triples as you can; and factor 24 into prime factors

- anonymous

im not understanding the triples thing.

- anonymous

24 is 2x2x2x3

- anonymous

exactly... there are 3 2s so they can come out

- anonymous

cuberoot(24) is 2 times cuberoot(3)

- anonymous

fifteen xs so you can pull 5 triples out (all of them)

- anonymous

so its 2x^5

- anonymous

on the outside?

- anonymous

or is the answer 2x^5 y^2?

- anonymous

so far, but dont forget the ys

- anonymous

yes... but there is something left under the radical

- anonymous

3?

- anonymous

yes

- anonymous

you got it.... for square roots you pull out pairs, for cube roots you pull out the triples

- anonymous

so the answer is 2x^5 y^2 cube root 3?

- 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

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

yes, that is the answer... dnag you are good
and let's see the fraction

- anonymous

thank you!

- anonymous

I'm not sure how to write it, but its cube root of 72x^15 y^13/64x^12 y^9

- anonymous

\[\sqrt[3]{72x^15y^13 \over 64x^12y^9}\]

- anonymous

simplify it first ... write the numbers as prime factors and cancel as much as you can

- anonymous

i need some help with that. I don't understand.

- anonymous

you can cancel 12 xs and 9 ys right off the bat

- anonymous

so is it cube root of 72x^3 y^4/64xy?

- anonymous

\[{x^{15} \over x^{12} } = x^3\]

- anonymous

15 - 12 leaves 3 xs, and they happened to be on top

- anonymous

so there are none left on the bottom, just the 64?

- anonymous

exactly... got lucky this time

- anonymous

with the problem I mean, you are doing great

- anonymous

thank you.

- anonymous

so then i factor 72 and 64 or?

- anonymous

now cancel out the common factors in 72 and 64

- anonymous

yes, thats how to do it

- anonymous

3x3x2x2x2?

- anonymous

thats 72 it seems

- anonymous

3x3x2x2x2 / 2x2x2x2x2x2
3x3 - - - - - - 2x2x2

- anonymous

2x2x3x3?

- anonymous

i know that last one is wrong.

- anonymous

keep dividing by 2 until you can't do it any more
then see if what is left is prime

- anonymous

idk

- anonymous

64 is 2x2x2x2x2x2

- anonymous

right so that is 3

- anonymous

because there are 3 sets right?

- anonymous

not yet, we are trying to simplify... cancelling what we can

- anonymous

oh

- anonymous

that 72 is still up there, I mean the 3x3x2x2x2

- anonymous

so 3x3x2x2x2 / 2x2x2x2x2x2 is
3x3 - - - - - - 2x2x2

- anonymous

cacelled out 2x2x2 from top and bottom

- anonymous

ok

- anonymous

so that leaves 9 x^3y^4 over 8 in the radical

- anonymous

what goes on the outside?

- anonymous

we havent pulled anything out yet, just simplified it
you pick the triples to pull out (since this is a cube root)

- anonymous

is it just one y on the outside?

- anonymous

\[\sqrt[3]{9x^3y^4 \over 8}\]

- anonymous

yes, pull 3ys out, one ouside and leaves one inside

- anonymous

\[3y^{3}\] and \[x^{3}\] on the outside

- anonymous

and then 9y/8 on the inside?

- 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

which does leave 9y inside ... and the cube root of 1/8 is 1/2

- anonymous

1/8xy on the outside

- anonymous

1/2 x y on the outside

- 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

so an x outside represents 3 inside, and so on

- anonymous

ok so then its just x^2 y^2 on the inside?

- anonymous

just a y on the inside?

- anonymous

I think the answer is:
\[{1 \over 2}xy \sqrt[3]{9y}\]

- anonymous

you are right.

- anonymous

you can check it by cubing what is on the left and multiplying that by what is under the radical

- anonymous

there are lots of shortcuts, but they always confuse me
I like doing things the long simple way

- anonymous

okay great. thank you so much for your help. that was a hard one.

- anonymous

only because of the typing... you knwo what you are doing
keep up the good work

- anonymous

thank you.

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