<s>[INST] Prove that if $abc\ne0$ and $ab+bc+ac=0$ then $a+b+c\ne0$ I tried to do proof  by contradiction, but problem is how to get from $ab+bc+ac$ to $a+b+c$
Assuming $a+b+c=0$
my approachs:


*

*Adding $ab+ac+bc=0$ and $a+b+c=0$ and try to factor

*Deriving $$a^2+ab+ac=0\\ac+bc+c^2=0\\ab+b^2+bc=0$$ and trying to derive something but nothing useful. 


Please just some hint will do the help to me 
 [/INST] If $ab+bc+ac=a+b+c=0$, then $a,b,c$ are the three roots of a cubic equation $X^3 - abc=0$.  But any nonzero number has two non-real cube roots, so $abc=0$—assuming that at least two of $a,b,c$ are real.
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