1. ## Demonstration and deduction

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

Deduct all resolutions of this equation in R :

2. Originally Posted by dhiab
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 $a^3 + b^3 + c^3 - 3abc = \frac12 \: (a+b+c) \: \left[(b-c)^2+(c-a)^2+(a-b)^2\right]$

3. [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 :

4. Originally Posted by running-gag
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 :

5. Originally Posted by dhiab
HELLO :
equality is correct : LOOCK THIS RESOLUTION

Deduct :

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