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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?

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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 $x_2 = \pm \frac{\sqrt{3}}{2}$.

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