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Math Help - factorization problem

  1. #1
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    factorization problem

    If  [(x/\sqrt{y})^3+(y/\sqrt{x})^3]^2+[(x/\sqrt{y})^3-(y/\sqrt{x})^3]^2=10(x^3+y^3)..............(1)

    and  (\frac{x^6-y^6}{6x^4y^4})^3=x^3+y^3.........(2)

    then calculate \frac{3x^3y^3}{y^3-x^3}


    Any suggestion of how to get to the solution will be appreciated.
    Note: the final answer not necessary has to be an integer(could be a simpler form)

    Thanks in advance
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: factorization problem

    Quote Originally Posted by rochosh View Post
    If  [(x/\sqrt{y})^3+(y/\sqrt{x})^3]^2+[(x/\sqrt{y})^3-(y/\sqrt{x})^3]^2=10(x^3+y^3)..............(1)

    and  (\frac{x^6-y^6}{6x^4y^4})^3=x^3+y^3.........(2)

    then calculate \frac{3x^3y^3}{y^3-x^3}


    Any suggestion of how to get to the solution will be appreciated.
    Note: the final answer not necessary has to be an integer(could be a simpler form)

    Thanks in advance
    please post your attempt. it is easier that way.
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  3. #3
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    Re: factorization problem

    Solving the first equation (1) further enough you'll get:
    (x^3+y^3)^2=8x^3y^3...........(I)

    Factorizing and reducing the second one (2):
    (x^3-y^3)^3(x^3+y^3)^2=(6x^4y^4)^3..........(II)

    Replacing (I) in (II):
    (x^3-y^3)^{3}8x^3y^3=(6x^4y^4)^3

    Solving this final equation in order to find the required one above and you'll get
    -1=\frac{3x^3y^3}{y^3-x^3}

    Reorganizing my attempt to post it, I casually found out the solution.
    In any case another way to solve this problem will be appreciated.
    Thanks for the remarking.
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Re: factorization problem

    Quote Originally Posted by rochosh View Post
    Solving the first equation (1) further enough you'll get:
    (x^3+y^3)^2=8x^3y^3...........(I)

    Factorizing and reducing the second one (2):
    (x^3-y^3)^3(x^3+y^3)^2=(6x^4y^4)^3..........(II)

    Replacing (I) in (II):
    (x^3-y^3)^{3}8x^3y^3=(6x^4y^4)^3

    Solving this final equation in order to find the required one above and you'll get
    -1=\frac{3x^3y^3}{y^3-x^3}

    Reorganizing my attempt to post it, I casually found out the solution.
    In any case another way to solve this problem will be appreciated.
    Thanks for the remarking.
    see i told you it works!
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