If a^3+b^3+c^3 = 3abc , find the value of a+b+c?

Answer 1

(1,1,1), (0,0,0), and (-1,-1,-1) are all solutions, so there isn't a unique value for a+b+c. 3, 0, and -3 at least, are possibilities.

Answer 2

a+b+c = 3, one solution at least.

Answer 3

abc

Answer 4

(a+b+c)^3=(a^3+b^3+c^3)+3a^2b+3a^2c+ 3b^2a+3b^2c+3c^2 a+3c^2b+6abc
(a+b+c)^3=3abc+3a^2b+3a^2c+3b^2a+ 3b^2c+3c^2a+3c^2b+6abc
=3abc+3a^2b+3a^2c+3b^2a+ 3b^2c+ 3c^2a+3c^2b+3abc+3abc
rearrange them
(a+b+c)^3=(3abc+3a^2b+3b^2a)+
(3abc+3b^2c+3c^2b) +(3abc+3a^2c+3c^2a)
=3ab(c+b+a)+3bc(a+b+c)+3ac(b+a+c)
=(a+b+c)(3ab+3bc+3ac)
(a+b+c)^2=(3ab+3bc+3ac)
=3(ab+bc+ac)
a+b+c = sqrt{3(ab+bc+ca)}