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Thread: Demostration of an equation condinioned by "a+b+c=0"

  1. #1
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    Demostration of an equation conditioned by "a+b+c=0"

    I wil appretiate any suggestion of how to demostrate
    this equation:

    if $\displaystyle a+b+c=0$
    then $\displaystyle (a^2+b^2+c^2)(a^3+b^3+c^3)/6=(a^5+b^5+c^5)/5$

    I could investigate for myself some features and
    found some interesting expressions, I list them below:

    $\displaystyle (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$................(1).
    $\displaystyle =a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc$.........(2).
    $\displaystyle =a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc$....(3).
    a mean they are all equal (1)=(2)=(3).

    and if $\displaystyle a+b+c=0$ then
    (4)......$\displaystyle 3abc=a^3+b^3+c^3$.........see exppretion 2.
    (5)......$\displaystyle -2(ab+bc+ac)=a^2+b^2+c^2$.........It is well known that$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$.

    I hope these ecuations above help.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: Demostration of an equation conditioned by "a+b+c=0"

    Quote Originally Posted by rochosh View Post
    I wil appretiate any suggestion of how to demostrate
    this equation:

    if $\displaystyle a+b+c=0$
    then $\displaystyle (a^2+b^2+c^2)(a^3+b^3+c^3)/6=(a^5+b^5+c^5)/5$



    I could investigate for myself some features and
    found some interesting expressions, I list them below:

    $\displaystyle (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$................(1).
    $\displaystyle =a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc$.........(2).
    $\displaystyle =a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc$....(3).
    a mean they are all equal (1)=(2)=(3).

    and if $\displaystyle a+b+c=0$ then
    (4)......$\displaystyle 3abc=a^3+b^3+c^3$.........see exppretion 2.
    (5)......$\displaystyle -2(ab+bc+ac)=a^2+b^2+c^2$.........It is well known that$\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$.

    I hope these ecuations above help.
    there's a very systematic way to do this problem.
    Let $\displaystyle a, b, c$ be the roots of $\displaystyle f(x)=x^3+ \alpha x - \beta=0$. Nothe that the coefficient of $\displaystyle x^2$ is $\displaystyle 0$ because $\displaystyle a+b+c=0$.Note that $\displaystyle \alpha = ab + bc+ca, \beta=abc$
    NOTATION: $\displaystyle \Sigma a= a+b+c, \Sigma a^2= a^2+b^2+c^2$,$\displaystyle \Sigma ab= ab+bc+ca$ and so on.


    1) $\displaystyle (\Sigma a)^2= \Sigma a^2 + 2 \Sigma ab \Rightarrow \boxed{ \Sigma a^2=-2 \Sigma ab=-2 \alpha}$

    2)$\displaystyle f(a)=f(b)=f(c)=0 \Rightarrow f(a)+f(b)+f(c)=0 \Rightarrow \boxed{\Sigma a^3= - \alpha (\Sigma a)+3 \beta = 3 \beta}$.

    3)define$\displaystyle h(x)=x^2 f(x)$.
    $\displaystyle \text{Then } h(a)=h(b)=h(c)=0 \Rightarrow h(a)+h(b)+h(c)=0 \Rightarrow \boxed{\Sigma a^5=- \alpha (\Sigma a^3) + \beta (\Sigma a^2)=-5 \alpha \beta}$.

    Using the boxed relations the required result is immediate.
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    Re: Demostration of an equation conditioned by "a+b+c=0"

    I have to say thank for the reasonable answer, but would you mind to write me back with any references about this topic(like book titles),
    I asking you this becouse I'd like to know the way you easily find out the solution.
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Re: Demostration of an equation conditioned by "a+b+c=0"

    Quote Originally Posted by rochosh View Post
    I have to say thank for the reasonable answer, but would you mind to write me back with any references about this topic(like book titles),
    I asking you this becouse I'd like to know the way you easily find out the solution.
    i found this technique in a math Olympiad book while i was still in high school. i don't a particular book which covers this technique. Its just a random weapon in my problem solving arsenal.
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