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Math Help - Cubed Root Question

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    Cubed Root Question

    In the following factoring question

    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 64x^3 is 4x, but does one do the cubed root of  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 27^6 is 3y^2, but I do not know specifically how the y^6 turns into y^2 (obviously the cube root of 27 is 3).

    Could some one please clarify and show me?
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  2. #2
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    Hello,



    Remember the rules of the powers :

    x^{yz}=(x^y)^z=(x^z)^y

    So here :
    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
    y^6 by itself, how would I know the correct order to factor it? I could do (y^{3*2}) = (y^3)^2 instead of the correct order  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
    y^6 by itself, how would I know the correct order to factor it? I could do (y^{3*2}) = (y^3)^2 instead of the correct order  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 (y^2)^3 or (y^3)^2, according to what you need
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  5. #5
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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
    y^6 by itself, how would I know the correct order to factor it? I could do (y^{3*2}) = (y^3)^2 instead of the correct order  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.

    (a^3-b^3)

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