Adding all of the equation, we'll get $a+b+c=0$. Note that \[ (-c)(a-b) = (a+b)(a-b) = a^2 - b^2 = -a \]\[ (-a)(b-c) = (b+c)(b-c) = b^2 - c^2 = -b \]\[ (-b)(c-a) = (a+c)(c-a) = c^2 - a^2 = -c \]Multiply three of these equations, and we'll get the desired result: $(a-b)(b-c)(c-a) = 1$.

Try to prove such numbers $a,b,c$ exist.

The system has only four solutions and they all turn out to be in $\mathbb{R}$. They are $(0,0,0)$ and the cyclic permutations of $\left(\frac{2}{3}\cos\frac{2\pi}{9}-\frac{2}{3}\cos\frac{4\pi}{9},\frac{2}{3}\cos\frac{4\pi}{9}-\frac{2}{3}\cos\frac{8\pi}{9},\frac{2}{3}\cos\frac{8\pi}{9}-\frac{2}{3}\cos\frac{2\pi}{9}\right)$ We discard the $(0,0,0)$ solution and noting $\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{8\pi}{9}=0$ gives: $(a-b)(a-c)(b-c)=-8\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=-(e^{2\pi i/9}+e^{-2\pi i/9})(e^{4\pi i/9}+e^{-4\pi i/9})(e^{8\pi i/9}+e^{-8\pi i/9})$ $=-(e^{-4\pi i/9}+e^{-2\pi i/9}+e^{2\pi i/3}+e^{8\pi i/9}+e^{-8\pi i/9}+e^{-2\pi i/3}+e^{2\pi i/9}+e^{4\pi i/9})$ $=-2\left(\cos\frac{4\pi}{9}+\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{2\pi}{3}\right)=1$

$\color{blue} \boxed{\textbf{SOLUTION}}$ Here, $(c+a)=b^{2}+a^{2}-a^{2}-c^{2}=b^{2}-c^{2}$ Similarly, $(a+b)=c^{2}-a^{2}$ $(b+c)=a^{2}-b^{2}$ So, $\prod (a+b)=\prod (c^{2}-a^{2})=\prod (a^{2}-b^{2})$ $\prod (a-b)=\frac {\prod (a^{2}-b^{2})}{\prod (a+b)}$ $=\frac {\prod (a^{2}-b^{2})}{\prod (a^{2}-b^{2})}=1$ Here is a case that if $a=-b$ or $b=-c$ or $c=-a$ then we can't divide but there is a contradiction, WLOG, $a=-b$ From the first equation we get, $(-b)^{2}-b=b^2 \implies b=0$ Contradiction! $\blacksquare$

mruczek wrote: Real numbers $a, b, c$ are not equal $0$ and are solution of the system: $\begin{cases} a^2 + a = b^2 \\ b^2 + b = c^2 \\ c^2 +c = a^2 \end{cases}$ Prove that $(a - b)(b - c)(c - a) = 1$. İf we sum the system of equations side by side then we get $a+b+c=0$ and if we sum the first and second equations we get $a^2=c^2+c$ $\rightarrow$ $c-a=\frac{-c}{a+c}=\frac{c}{b}$ in the same manner $a-b=\frac{a}{c}$ and $b-c=\frac{b}{a}$ from there we get $(a-b)(b-c)(c-a)=$$\frac{c}{b}$$\frac{a}{c}$$\frac{b}{a}$$=1$