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Math Help - Need to facto this.

  1. #1
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    Need to facto this.

    factoRRRR this...and I don't know where to start

    y^6-1
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  2. #2
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    Quote Originally Posted by Alienis Back View Post
    factoRRRR this...and I don't know where to start

    y^6-1
    = (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....

    And you should know how to factorise the sum and difference of two perfect cubes.
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  3. #3
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    Do you mean this??

    = (y^2)^3 - 1^3

    Where a=y^2 and b=1

    a^3-b^3=(a-b)(a^2+ab+b^2)

    Then we have...

    (y^2)^3-1^3=(y^2-1) [(y^2)^2+y^2+1)]

    (y^2)^3-1^3=(y^2-1) (y^4+y^2+1)

    ...But I don't think this is the answer...mmm...I really don't
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  4. #4
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    Quote Originally Posted by Alienis Back View Post
    Do you mean this??

    = (y^2)^3 - 1^3

    Where a=y^2 and b=1

    a^3-b^3=(a-b)(a^2+ab+b^2)

    [snip]
    How can you possibly think I mean that when I clearly wrote this:

    = (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....

    And you should know how to factorise the sum and difference of two perfect cubes.
    Now all you have to do is factorise each of y^3 - 1 = y^3 - 1^3 and y^3 + 1 = y^3 + 1^3.
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