Check out some similar questions!

peri42 · Aug 22, 2009, 04:24 PM

Simplify with positive exponents..
5x^3y^6 under radical sign
divided by 3a^3

Leidenschaftlich für Wahr · Aug 22, 2009, 04:27 PM

Do your own homework...

galactus · Aug 22, 2009, 05:27 PM

Is this it: $\frac{\sqrt{5x^{3}y^{6}}}{3a^{3}}$ ?

If so, there is not much to be done---except look at the y term.

peri42 · Aug 22, 2009, 09:21 PM

you so would it be y^3 * y^3?

morgaine300 · Aug 22, 2009, 11:24 PM

Yes, and that is $(y^3)^2$ , which is under a radical. You can do something with $sqrt{(y^3)^2}$
Think about it.

peri42 · Aug 23, 2009, 01:52 PM

2 squared radical y?I'm not sure

peri42 · Aug 23, 2009, 02:31 PM

no wudnt it just be y^3?

morgaine300 · Aug 24, 2009, 12:53 AM

Originally Posted by peri42

no wudnt it just be y^3?

Yup. :-) The square root of a squared number is just that number.

Remember that you have to pull that out from under the radical since you took the square root of it. But I don't see anything further you can do to the rest of it.

peri42 · Aug 24, 2009, 09:07 AM

is xy^2 5 under radical sign over 3a^3 right?

galactus · Aug 24, 2009, 11:35 AM

You could write it as $\frac{\sqrt{5x^{3}}y^{3}}{3a^{3}}$

Remember that $\sqrt{y^{6}}=(y^{6})^{\frac{1}{2}}=y^{3}$

Not much can be done with the x^3 inside the radical. That is the same as $\sqrt{x^{3}}=x^{\frac{3}{2}}$ , so just leave it alone.

Actually, if we were to be purists, the actual form would be:

$\frac{\sqrt{5x^{3}}\cdot |y^{3}|}{3a^{3}}$

But I do not believe it is necessary to introduce the absolute value, though technically, it belongs there.

peri42 · Aug 26, 2009, 03:11 PM

OK thnks 4 your help =)