Question

OK can someone help with this problem ill post the link any help is appreciated

Answer 1

whats the problem, thats a perfectly reasonable quotient

Answer 2

I have to simplify it as much as possible

Answer 3

did you learn conjugate and other stuff ?

Answer 4

yes but its a little fuzzy it was last year so i don't quite remember

Answer 5

you have to get rid of the cube root if i am not mistaken

Answer 6

well, third root of 27 is 3, so you can pull the 27 out and just times the bottom by 3, third root of 8 is 2, so you can do the same on the top

and you can multiply the top and bottom by whats on the bottom, twice, to get rid of the cube root at the bottom

Answer 7

then simply slip all the top bit under the square root sign, since its all cube rooted.

Answer 8

There isn't much you can really do unless you assume x and y are positive. Otherwise, this is about as simplified as I can make it:
(2/3) * ((x^18 * y^10) / ( x^6 * y))

Assuming x and y are positive you could do this:

Assume z equals your little equation.

Take both sides to the power of three:

z^3 = (8* x^18 * y^10) / (27 * x^6 * y)

Now simplify the right side:

- Bring out the (8/27)
- Simplify like variables

z^3 = (8/27)*(x^12 * y^9)

Find the cube root of both sides:

z = ((8/27)*(x^12 * y^9))^(1/3)

Now, apply above suggestions:

- Bring the (8/27) out of the cube root

(2/3)*((x^12 * y^9))^(1/3)]

(2/3) (x^4 * y^3)

That is as far as I can simplify it (again, assuming x and y are positive)