# Find the critical points of the following system

I want to find the critical points of the following three-dimensional system:

\begin{align} \dot{x_1} &= x_1 - x_1x_2 - x_2^3 + x_3(x_1^2 + x_2^2 - 1 - x_1 + x_1x_2 + x_2^3)\\ \dot{x_2} &= x_1 - x_3(x_1 - x_2 + 2x_1x_2)\\ \dot{x_3} &= (x_3 - 1)(x_3 + 2x_3x_2^2 + x_3^3) \end{align} I know that the critical points of the system can be found by solving \begin{align} x_1 - x_1x_2 - x_2^3 + x_3(x_1^2 + x_2^2 - 1 - x_1 + x_1x_2 + x_2^3)= 0\\ x_1 - x_3(x_1 - x_2 + 2x_1x_2) = 0\\ (x_3 - 1)(x_3 + 2x_3x_2^2 + x_3^3) = 0 \end{align} for $$x$$, so that's what I've tried to do. From the second equation I find that $$x_3 = \dfrac{x_1}{x_1 - x_2+2x_1x_2}$$ From the third equation I find that $$x_2 = \sqrt{-1/2 - x_3^2/2}$$ or $$x_3 = 1$$. Unfortunately, I don't really know how to proceed from here. If I plug in $$x_3$$ in $$x_2$$ I don't get anywhere. Furthermore, I don't know how to find $$x_1$$.

Question: How do I find the critical points of the system described above?

I would start with the 3rd equation $$(x_3-1)x_3(x_3^2+2x_2^2 + 1)=0.$$ Since the third factor is always positive, $$x_3= 0$$ or $$x_3=1$$.
Case A: $$x_3=0$$. Substituting $$x_3=0$$ in equation 2 gives $$x_1=0$$. Then equation 1 reduces to $$-x_2^3 =0$$, so $$x_2=0$$.
Case B: $$x_3=1$$. Substituting $$x_3=1$$ in equation 1 gives $$x_1^2+x_2^2=1$$. Moreover, equation 2 reduces to $$x_2(1-2x_1)=0$$. From the latter it follows that $$x_2=0$$ or $$x_1= 1/2$$.
Case B.1 $$x_3=1$$, $$x_2=0$$. Equation 1 simplifies to $$x_1^2=1$$, so $$x_1 = \pm 1$$.
Case B.2 $$x_3=1$$, $$x_1=1/2$$. Equation 1 simplifies to $$x_2^2=3/4$$, so $$x_2 = \pm \frac{\sqrt{3}}{2}$$.