a,b,c is the reals numbers , Prove this equality :
Deduct all resolutions of this equation in R :
[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:
$\displaystyle (b-c)^2= b^2- 2bc+ c^2$
$\displaystyle (c-a)^2= c^2- 2ac+ a^2$
$\displaystyle (a-b)^2= a^2- 2ab+ b^2$
so $\displaystyle (b-c)^2+ (c-a)^2+ (a-b)^2= 2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc)$
Now multiply that by a+ b+ c:
$\displaystyle a(2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc))= 2a^3+ 2ab^2+ 2ac^2- 2(a^2b+ a^2c+ abc)$
$\displaystyle b(2a^2+ 2b^2+ 2c^2- 2(ab+ac+ bc))= 2a^2b+ 2b^3+ 2bc^2- 2(ab^2+ abc+ 2bc^2)$
[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 "$\displaystyle 2ab^2$" term in the first equation is canceled by the "$\displaystyle -2ab^2$" term in the second equation, etc.
Deduct all resolutions of this equation in R :