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Thread: Need to facto this.

  1. #1
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    Need to facto this.

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

    $\displaystyle y^6-1$
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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

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

    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??

    $\displaystyle = (y^2)^3 - 1^3$

    Where $\displaystyle a=y^2$ and $\displaystyle b=1$

    $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

    Then we have...

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

    $\displaystyle (y^2)^3-1^3=(y^2-1) (y^4+y^2+1)$

    ...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??

    $\displaystyle = (y^2)^3 - 1^3$

    Where $\displaystyle a=y^2$ and $\displaystyle b=1$

    $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$

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

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

    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 $\displaystyle y^3 - 1 = y^3 - 1^3$ and $\displaystyle y^3 + 1 = y^3 + 1^3$.
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