There are a couple of problems I can't understand as usual. I tried to solve them, therefore I will give what I believe is the solution next to it.

(1). $\sqrt[3]{-125x^3z^1^5}$ Solution: $-5x&z^3$

(2). $\sqrt[3]{\frac{-125x^5y^4z^7}{8x^2y^1^0z^4}}$ This one is extremely of what its really supposed to be, I can't understand how to take the powers very well. Solution: $\frac{-5xyz^2\sqrt[3]{x^2yz}}{2y^3z\sqrt[3]{x^2yz}}$

(3). $\sqrt[3]{\frac{8}{9}(\frac{3a^2b}{a^5b4})^5}$

Solution:I can't simplify the fraction 8/9 I'm asuming $\frac{2}{\sqrt[3]{9}}$

Originally Posted by vaironxxrd
There are a couple of problems I can't understand as usual. I tried to solve them, therefore I will give what I believe is the solution next to it.

(1). $\sqrt[3]{-125x^3z^1^5}$ Solution: $-5x&z^3$

(2). $\sqrt[3]{\frac{-125x^5y^4z^7}{8x^2y^1^0z^4}}$ This one is extremely of what its really supposed to be, I can't understand how to take the powers very well. Solution: $\frac{-5xyz^2\sqrt[3]{x^2yz}}{2y^3z\sqrt[3]{x^2yz}}$

(3). $\sqrt[3]{\frac{8}{9}(\frac{3a^2b}{a^5b4})^5}$

Solution:I can't simplify the fraction 8/9 I'm assuming $\frac{2}{\sqrt[3]{9}}$
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(1): How did you get a power of three on the z ?

(2): In your result, reduce the fraction! The radicals are identical, therefore they cancel. Outside the radical, other quantities cancel.

(3): Rationalize the denominator.

What happened to the factor to the power 5 ?

Originally Posted by SammyS
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What happened to the factor to the power 5 ?
(1): How did you get a power of three on the z ?
3 groups of 5

(2): In your result, reduce the fraction! The radicals are identical, therefore they cancel. Outside the radical, other quantities cancel.
$\frac{-5z}{2y^2z}$

(3): Rationalize the denominator.
Ill look on how to do that.

Originally Posted by vaironxxrd
(1): How did you get a power of three on the z ?
3 groups of 5
Last time I checked $\dfrac{15}{3}$ wasn't 3 . You've done it fine for each term so I suspect it's an honest mistake rather than not knowing

(2): In your result, reduce the fraction! The radicals are identical, therefore they cancel. Outside the radical, other quantities cancel.
$\frac{-5z}{2y^2z}$
Almost but you've not got rid of z on the denominator.

Edit: Any luck on rationalising the denominator? The idea is to get rid of that $\sqrt[3]{9}$ by multiplying top and bottom by something that would make it rational. In this case you'll want to get 9 on the denominator

Originally Posted by vaironxxrd
...

(3). $\sqrt[3]{\frac{8}{9}(\frac{3a^2b}{a^5b4})^5}$
The more important question for (3) is What happened to this $\left(\frac{3a^2b}{a^5b^4}\right)^5$ under the radical?

The more important question for (3) is What happened to this $\left(\frac{3a^2b}{a^5b^4}\right)^5$ under the radical?
$\left(\frac{3^5a^1^0b^5}{a^2^5b^2^0}\right)$