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Thread: Simplifying Radicals

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    Senior Member vaironxxrd's Avatar
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    Simplifying Radicals

    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 $\displaystyle \frac{2}{\sqrt[3]{9}}$
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    Re: Simplifying Radicals

    Quote Originally Posted by vaironxxrd View Post
    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 ?
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    Senior Member vaironxxrd's Avatar
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    Re: Simplifying Radicals

    Quote Originally Posted by SammyS View Post
    .

    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.
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    Re: Simplifying Radicals

    Quote Originally Posted by vaironxxrd View Post
    (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 . 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.
    $\displaystyle \frac{-5z}{2y^2z}$
    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
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    Re: Simplifying Radicals

    Quote Originally Posted by vaironxxrd View Post
    ...

    (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?
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    Re: Simplifying Radicals

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