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Math Help - Ploynomial...

  1. #1
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    Ploynomial...

    Not sure if im going mad but struggling with this
    (a) Write down the polynomial P in x, y such that:
    x^6 y^6 = (x^2 y^2)*P.

    Many thanks in advance.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Hint :

    Denote t=x^2 and a=y^2 then,

    x^6-y^6=t^3-a^3

    so,

    a is a root of q(t)=t^3-a^3

    and you can decompose it using Ruffini's rule.


    Fernando Revilla
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  3. #3
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    Hello, breitling!

    \text{Write down the polynomial }P\text{ in }x, y\text{ such that:}

    . . x^6 - y^6 \:=\: (x^2 - y^2)\cdot P

    Don't you recognize that difference of cubes?

    . . . . a^3 - b^3 \:=\:(a-b)(a^2 + ab + b^2)



    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~



    The very least you can do is some Long Division . . .


    . . \begin{array}{cccccccccc} &&&& x^4 & + & x^2y^2 &+& y^4 \\ && -- & -- & -- & -- & -- & -- & -- \\ x^2-y^2 & | & x^6 &&&&& - & y^6 \\ && x^6 &-& x^4y^2 \\ && -- & -- & -- \\ &&&& x^4y^2 \\ &&&& x^4y^2 &-& x^2y^4 \\ &&&& -- & -- & -- \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& x^2y^4 &-& y^6 \\ &&&&&& -- & -- & --  \end{array}

    Last edited by Soroban; January 25th 2011 at 06:23 AM.
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  4. #4
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    A better solution:

    \displaystyle x^6 - y^6 = (x^2)^3 - (y^2)^3

    \displaystyle = (x^2 - y^2)[(x^2)^2 + x^2y^2 + (y^2)^2] by the Difference of Two Cubes rule...

    \displaystyle = (x^2 - y^2)(x^4 + x^2y^2 + y^4).


    So what is \displaystyle P?
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  5. #5
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    The question actually asked x^6-y^6=(x^2+y^2)*p got the sign wrong.
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  6. #6
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    Quote Originally Posted by breitling View Post
    The question actually asked x^6-y^6=(x^2+y^2)*p got the sign wrong.
    Then do exactly what people suggested you do before!
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