<s>[INST] Solution of system of equations involving $x_{1},x_{2},x_{3}$ 
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$

Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_{3}-x_{1})$
Could some help me to find $x_{1},x_{2},x_{3}$, Thanks
 [/INST] I would start with solving $$ax^2_{1}+bx_{1}+c=x_{2}$$
Via the quadratic formula,
$$x_1=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}$$
Repeating the same process, we have
$$x_2=\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}$$
and
$$x_3=\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}$$
Hence,
\begin{align}
x_1&=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}\right)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}\right)}}{2a}\right)}}{2a}\\
\end{align}
in which $x_1$ can be painfully solved for
to get
$$x_1=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
Then just repeat for $x_2$ and $x_3$ for which you will get
$$x_1=x_2=x_3=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
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