It's easy to see that if $a+b+c=0$ then all the identities are true (just write $a=-b-c$ and plug this in). Now, if the three equalities are true, then w.l.o.g. $a \ne b$. So $b^2+bc=a^2+ac$ which rewrites as $0=a^2-b^2+ac-bc=(a-b)(a+b+c)$. Since we assumed $a \ne b$, this implies $a+b+c=0$ and we are done.

$a+b=–c$ $a^2+b^2+2ab=c^2$ Plug it in And get $c^2–ab=c^2+ac+a^2$ $\implies –a=(b+c)$. This is true and will give $a+b+c=0$ For take $P(a,b)=P(b,c)$ $P(b,c)=P(a,c)$ and again $P(a,c)=P(a,b)$ Equating all we get $a+b+c=0$ Note a similar problem appeared few years back

Extremely obvious after noting $ a\neq b$

a*a+a*b+b*b-b*b-b*c-c*c=0 a*a-c*c+a*b-b*c=0 (a-c)*(a+c)+b*(a-c)=0 (a-c)*(a+b+c)=0 a+b+c=0

Vibing yeah!/: just hot a+b+c=0 by using eqn we just see Either a=b=c Or a+b+c=0 But a=b=c dont work thus we are done