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Math Help - Demonstration and deduction

  1. #1
    Super Member dhiab's Avatar
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    Demonstration and deduction

    a,b,c is the reals numbers , Prove this equality :


    Deduct all resolutions of this equation in R :

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  2. #2
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    Quote Originally Posted by dhiab View Post
    a,b,c is the reals numbers , Prove this equality :

    This is wrong. If you take a=b=c you get 3a^3 on the left side and 0 on the right side.

    I think the correct equality is a^3 + b^3 + c^3 - 3abc = \frac12 \: (a+b+c) \: \left[(b-c)^2+(c-a)^2+(a-b)^2\right]
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  3. #3
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    [QUOTE=dhiab;332083]a,b,c is the reals numbers , Prove this equality :


    Looks to me like the best thing to do is just go ahead and multiply out the right side:
    (b-c)^2= b^2- 2bc+ c^2
    (c-a)^2= c^2- 2ac+ a^2
    (a-b)^2= a^2- 2ab+ b^2
    so (b-c)^2+ (c-a)^2+ (a-b)^2= 2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc)
    Now multiply that by a+ b+ c:
    a(2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc))= 2a^3+ 2ab^2+ 2ac^2- 2(a^2b+ a^2c+ abc)
    b(2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc))= 2a^2b+ 2b^3+ 2bc^2- 2(ab^2+ abc+ 2bc^2)
    [tex]c(2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc))= 2a^2c+ 2b^2c+ 2c^3- 2(abc+ ac^2+ bc^2)

    Now note that the " 2ab^2" term in the first equation is canceled by the " -2ab^2" term in the second equation, etc.

    Deduct all resolutions of this equation in R :

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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by running-gag View Post
    This is wrong. If you take a=b=c you get 3a^3 on the left side and 0 on the right side.

    I think the correct equality is a^3 + b^3 + c^3 - 3abc = \frac12 \: (a+b+c) \: \left[(b-c)^2+(c-a)^2+(a-b)^2\right]
    HELLO :
    equality is correct : LOOCK THIS RESOLUTION




    Deduct :


    Conclusion :
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  5. #5
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    Quote Originally Posted by dhiab View Post
    HELLO :
    equality is correct : LOOCK THIS RESOLUTION




    Deduct :


    Conclusion :
    What have become the 3 factors equal to -2abc ?
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