Be careful, $a,b,c\ne0$

$\begin{cases} a^2 +a = b^2 \\ b^2 +b =c^2 \\ c^2 +c =a^2 \end{cases} \implies \begin{cases} (a-b)(a+b)= -a \\ (b-c)(b+c) = -b \\ (c-a)(c+a) = -c \end{cases}$ Note that $a+b+c=0$ ( on adding the equations above). Multiplying the Right hand side equation $\begin{aligned} &(a-b)(b-c)(c-a) = \dfrac{-(abc)}{(a+b)(b+c)(c+a)} \\& (a-b)(b-c)(c-a) = \dfrac{-(abc)}{(-c)(-b)(-a)} \phantom{ccccc} {\color{#3D99F6} a+b+c =0 } \\& (a-b)(b-c)(c-a) =\boxed{1}\end{aligned}$

$\begin{cases} a^2 + a = b^2 & ...(1) \\ b^2 + b = c^2 & ...(2) \\ c^2 + c = a^2 & ...(3) \end{cases}$

$\begin{aligned} (1)+(2)+(3): \quad a^2+b^2+c^2 +a+b+c & = a^2+b^2+c^2 \\ \implies a+b+c & = 0 \end{aligned}$

From (1):

$\begin{aligned} a^2 + a & = b^2 \\ a^2 - b^2 & = - a \\ (a-b)(a+b) & = - a \\ (a-b) & = \frac {-a}{\color{#3D99F6}a+b} = \frac a{\color{#3D99F6}c} & \small \color{#3D99F6} \text{Note that }a+b+c =0 \end{aligned}$

Similarly, $b-c = \dfrac ba$ and $c-a=\dfrac cb$ and $(a-b)(b-c)(c-a) = \dfrac ac \times \dfrac ba \times \dfrac cb = \boxed{1}$ .