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Dlakt · Jan 13, 2010, 11:00 AM

algebra- I have to combine radicals assuming all variables are in nonnegative form

-2 (^3sqrt125x^3y^4)+(3y^2^3sqrt8x^3)

galactus · Jan 13, 2010, 11:18 AM

Please tidy up your expression. I can not tell what that means. Please use proper grouping symbols.

The ^3 by itself?

Is it $-2(\sqrt{125x^{3}y^{4}})+(3y^{{2}^{3}}\sqrt{8x^{3}} )$

As is, that is what you have written.

Dlakt · Jan 13, 2010, 12:39 PM

that is almost correct-
(-2)^3sqrt 125^3y^4+3y^2^3sqrt8x^3

so negative 2( sqrt 125x^3y^4) variable of sqrt is 3
+3y^2 (sqrt 8x^3) variable to sqrt is another 3

I am sorry I just started algebra 2 and very confused with the symbols this is the best I can do.

galactus · Jan 13, 2010, 01:07 PM

Variable of sqrt is 3? I am sorry, I do not know what you're getting at for sure without guessing.

What is y^2^3? That means $y^{2^{3}}=y^{8}$

I would like to help, but without something definitive I can only guess.

$-2(\sqrt[3]{125x^{3}y^{3}})+(3y^{2}\sqrt[3]{8x^{3}})$

If I am understanding correctly, it is a cube root. This cancels the cubes.

$\sqrt[3]{125x^{3}y^{3}}=(5^{3}x^{3}y^{3})^{\frac{1}{3}}=5x y$

Dlakt · Jan 13, 2010, 01:08 PM

3's are on the outside of the sqrt symbol /3's are variables so it is not 125x^3 its 3 sqrt 125 if that helps.

Dlakt

Dlakt · Jan 13, 2010, 01:08 PM

3's are on the outside of the sqrt symbol /3's are variables so it is not 125x^3 its 3 sqrt 125 if that helps.

Dlakt

Dlakt · Jan 13, 2010, 02:30 PM

Yes you wrote it correctly so is my answer 5xy?

Dlakt · Jan 13, 2010, 03:08 PM

what about the other part of the problem (3y^2 cubed) sqrt 8x^3

Dlakt · Jan 13, 2010, 03:19 PM

OK could you help me with this problem then ?

multiply and simplify asssume all variable represent nonnegative numbers

5sqrtA(3sqrtB-5) has to be in positive form

I do not have math symbols on my computer

galactus · Jan 13, 2010, 04:03 PM

I do not have math symbols on my computer

I don't either. That is called LaTeX. It is a code that enables one to display in nice math format. Click on 'quote' to see what code I used to make it display that way.

As for $\sqrt[3]{8x^{3}}=\sqrt[3]{2^{3}x^{3}}$

This is a basic exercise in learning the exponent laws.

A cube root cancels out the cubes. After all, $(a^{m})^{n}=a^{mn}$

A cube root is the 1/3 power. What is 1/3 times 3? See why they cancel now?

$\sqrt[3]{2^{3}x^{3}}=(2^{3}x^{3})^{\frac{1}{3}}=2x$

Dlakt · Jan 13, 2010, 04:23 PM

I really really thank you for your help, but one more thing you forgot the part of the problem the problem is (-2)sqrt125 x^3 variable is 3 +3y^2 sqrt8x^3 variable of 8x is 3 on your answer above you forgot the +3y^2

I am following so far, but is it 2 separate answers or do you combine the answers. I had asked you another question but you are trying to answer this one and that's great, I will figure out the other problem trying to work on it on my own.

Just not sure what the final answers are that you are giving me to my problem sorry I am confused. Thank you