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Thread: Cubed Root Question

  1. #1
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    Cubed Root Question

    In the following factoring question

    $\displaystyle 64x^3 - 27y^6$

    (a - b) (a^2 + ab + b^2) is used

    However, I am stuck right off the bat. The cube root of $\displaystyle 64x^3$ is 4x, but does one do the cubed root of $\displaystyle 27y^6$. Its obviously a difference of perfect cubes because x^3 and y^6 are both multiples of 3. The book tells me the cubed root of $\displaystyle 27^6$ is $\displaystyle 3y^2$, but I do not know specifically how the $\displaystyle y^6 $turns into $\displaystyle y^2$ (obviously the cube root of 27 is 3).

    Could some one please clarify and show me?
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    Hello,



    Remember the rules of the powers :

    $\displaystyle x^{yz}=(x^y)^z=(x^z)^y$

    So here :
    $\displaystyle y^6=y^{2 \cdot 3}=(y^2)^3$

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    For a better understanding lets take this same term but out of the cubed root application.

    If I was to factor
    $\displaystyle y^6$ by itself, how would I know the correct order to factor it? I could do $\displaystyle (y^{3*2}) = (y^3)^2$ instead of the correct order $\displaystyle y^{2*3} = (y^2)^3$

    Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

    Thanks
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    Quote Originally Posted by allyourbass2212 View Post
    For a better understanding lets take this same term but out of the cubed root application.

    If I was to factor
    $\displaystyle y^6$ by itself, how would I know the correct order to factor it? I could do $\displaystyle (y^{3*2}) = (y^3)^2$ instead of the correct order $\displaystyle y^{2*3} = (y^2)^3$

    Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

    Thanks
    Both are correct

    So you can take $\displaystyle (y^2)^3$ or $\displaystyle (y^3)^2$, according to what you need
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    Quote Originally Posted by allyourbass2212 View Post
    For a better understanding lets take this same term but out of the cubed root application.

    If I was to factor
    $\displaystyle y^6$ by itself, how would I know the correct order to factor it? I could do $\displaystyle (y^{3*2}) = (y^3)^2$ instead of the correct order $\displaystyle y^{2*3} = (y^2)^3$

    Or is it just as simple as writing out the multiples in ascending order to get the correct answer?

    Thanks
    The idea here is to have a base raised to the 3rd power. That's what your rule requires: the difference of two cubes.

    $\displaystyle (a^3-b^3)$

    If you used $\displaystyle (y^3)^2$ , you wouldn't have a base cubed. You would have a base squared.
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