Question

Someone help me with dividing cube roots--i can display it better in equation. Someone help me with dividing cube roots--i can display it better in equation. @Mathematics

Answer 1

\[\sqrt[3]{x^3y^2z^7}\div \sqrt[3]{xy^4}\]

Answer 2

\[\sqrt[3]{x^3y^2z^7}\div \sqrt[3]{xy^4}=\frac{\sqrt[3]{x^3y^2z^7}}{\sqrt[3]{xy^4}}\]\[=\sqrt[3]{\frac{x^3y^2z^7}{xy^4}}=\sqrt[3]{\frac{x^2z^7}{y^2}}\]

Answer 3

this can be simplified further if you wish.

Answer 4

Yes, could you try that? I found the answer and it's actually:\[z^2\sqrt[3]{x^2yz}/y\]

Answer 5

\[=\sqrt[3]{\frac{x^2z^7}{y^2}}=\sqrt[3]{\frac{x^2z^6zy}{y^3}}=\frac{z^2\sqrt[3]{x^2yz}}{y}\]

Answer 6

the steps I performed were:
1. z^7 = z * z^6
2. 1/y^2 = y/y^3
3. cube root of z^6 = z^2
4. cube root of 1/y^3 = 1/y

do you follow?

Answer 7

Yes, sorry i didnt reply sooner. I hope you're on here often! :)

Answer 8

Thanks, do you think you can help me with domain? It might be a little hard to explain but i'll try.

Answer 9

I must find the domains of f + g and f-g
f(x)=-2x^(2/3) g(x)=7x^(2/3)

Answer 10

So basically i have to know the domains of 5x^(2/3) and -9^(2/3)
:) think you can help?