Equilibrium points of $\dot x(t)=-2\cdot x^3(t)$

Question

The following differential equation is given: $$ \dot x(t)=-2\cdot x^3(t)\qquad x(t)\in\mathbb R $$ I am asked to find the equilibrium points of the system. By definition, the equilibrium points are: $$ 0=-2\cdot x_e^3\Rightarrow x^3_e=0\Rightarrow x_e=0 $$ Therefore my answer is: "there is one equilibrium point, $x_e=0$". However, the answer to the question explicitly takes points off for mentioning just one equilibrium point. It says there are three equilibria: $x_{e_1}=0,x_{e_2}=0,x_{e_3}=0$. I don't understand the need for this redundancy, since in the end it's the same point. Am I missing something subtle here?