Question

given a+b+c=0, prove the equality a^3+b^3+c^3=3abc

Answer 1

Given a+b+c=0, prove the equality \[a^3+b^3+c^3=3abc \]

Answer 2

If you start with a + b + c = 0, and then cube both sides, you have

(a + b + c)^3 = 0^3

Which expands to

a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3 = 0

Then isolate a^3 + b^3 + c^3:

a^3 + b^3 + c^3 = -3a^2b - 3a^2c - 3ab^2 - 6abc - 3ac^2 - 3b^2c - 3bc^2

Then simplify the right side:
a^3 + b^3 c^3 = -3(a^2b + a^2c + ab^2 +2abc + ac^2 + b^2c + bc^2)

That's as far as I am able to get with this.
If you can prove
-3(a^2b + a^2c + ab^2 +2abc + ac^2 + b^2c + bc^2) = 3abc, then you will have done it.

Answer 3

Wow, thank you so much

Answer 4

i think this might work, not sure \[a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc).
\]

Answer 5

@satellite73 I will try your method too, thanks for your input!