• Oct 23rd 2011, 12:27 PM
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).$\displaystyle \sqrt[3]{-125x^3z^1^5}$ Solution:$\displaystyle -5x&z^3$

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

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

Solution:I can't simplify the fraction 8/9 I'm asuming $\displaystyle \frac{2}{\sqrt[3]{9}}$
• Oct 23rd 2011, 12:35 PM
SammyS
Quote:

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

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

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

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

.
(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 ?
• Oct 23rd 2011, 01:18 PM
vaironxxrd
Quote:

Originally Posted by SammyS
.

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.
$\displaystyle \frac{-5z}{2y^2z}$

(3): Rationalize the denominator.
Ill look on how to do that.
• Oct 23rd 2011, 01:24 PM
e^(i*pi)
Quote:

Originally Posted by vaironxxrd
(1): How did you get a power of three on the z ?
3 groups of 5

Last time I checked $\displaystyle \dfrac{15}{3}$ wasn't 3 (Wink). You've done it fine for each term so I suspect it's an honest mistake rather than not knowing

Quote:

(2): In your result, reduce the fraction! The radicals are identical, therefore they cancel. Outside the radical, other quantities cancel.
$\displaystyle \frac{-5z}{2y^2z}$
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 $\displaystyle \sqrt[3]{9}$ by multiplying top and bottom by something that would make it rational. In this case you'll want to get 9 on the denominator
• Oct 23rd 2011, 01:37 PM
SammyS
Quote:

Originally Posted by vaironxxrd
...

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

The more important question for (3) is What happened to this $\displaystyle \left(\frac{3a^2b}{a^5b^4}\right)^5$ under the radical?
• Oct 23rd 2011, 02:28 PM
vaironxxrd
The more important question for (3) is What happened to this $\displaystyle \left(\frac{3a^2b}{a^5b^4}\right)^5$ under the radical?
$\displaystyle \left(\frac{3^5a^1^0b^5}{a^2^5b^2^0}\right)$