$a+b+c=0$ $a^2-b^2=c=-a-b \to (a+b)(a-b+1)=0$ Case 1: $a+b=0 \to c=0, \to abc=0$ Case 2: $b=a+1$ $ c=-(2a+1)$ $a^2+a+1=(2a+1)^2 \to a=0,-1$ So $abc=0$ Btw, all solutions are $(0,0,0),(-1,0,1),(0,1,-1),(1,-1,0)$

From the given condition, we can have: $ a+b+c=0 $ and $a^3+b^3+c^3=0$. Since \[ a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca), \]we get $abc=0$.

bigant146 wrote: Let $a$, $b$, $c$ be real numbers such that $a^2+b=c^2$, $b^2+c=a^2$, $c^2+a=b^2$. Find all possible values of $abc$. $$ a+b+c=b^2-c^2+c^2-a^2+a^2-b^2=0 $$$$a^3+b^3+c^3=a^2(b^2-c^2)+b^2(c^2-a^2)+c^2(a^2-b^2)=0$$$$ a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\implies abc=0$$

We claim that $abc$ must equal $\boxed0$. Adding the equations gives $a+b+c=0$. Then $c^2=a^2+b=a^2-a-c$, so $(a+c)(a-c-1)=0$. If $a+c=0$, then $b=0$ and thus $abc=0$. If $a=c+1$, then we have $c^2+c+1=c^2+a=b^2=(a+c)^2=(2c+1)^2$, so $c\in\{0,-1\}$. If $c=0$ then $abc=0$, where as if $c=-1$ then $a=0$ and the same result holds.

its copy from polish 2018 https://artofproblemsolving.com/community/c6h1632593p10254701

Simple simplifing Gives $ a= b^2-c^2= (c-b)\cdot a$ $b=c^2-a^2=(a-c)\cdot b$ $c=a^2-b^2=(b-a)\cdot c$ a,b,c $\rightarrow$ (0,0,0) and permutations of( 0,-1,1) thus $abc=0$

a²+b=c² b²+c=a² c²+a=b² We need abc. Okay, first find; b=c²-a² c=a²-b² a=b²-c² a+b+c=c²-a²+a²-b²+b²-c² a+b+c=0 After that we will look the cases. There are two cases, such as; i) a=b=c=0 ii) a=-b in that time c=0 b=-c in that time a=0 c=-a in that time b=0 All the solutions, there is a number that's equal to zero. So multiple will be 0 all the times. abc=0 <_>