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

**rochosh** · July 3rd 2011, 03:49 PM

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

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

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

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

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

I hope these ecuations above help.

**abhishekkgp** · July 3rd 2011, 05:34 PM

Originally Posted by rochosh

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

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

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

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

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

I hope these ecuations above help.

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

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

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

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

Using the boxed relations the required result is immediate.

**rochosh** · July 4th 2011, 12:50 PM

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.

**abhishekkgp** · July 4th 2011, 05:13 PM

Originally Posted by rochosh

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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