how to solve this differential equation with $x^3$?

Your differential equation is of the form $\frac{dx}{dt} = f(x) g(t)$. This type of equation is called a separable differential equation because you can solve it by separating the "$x$" part and the "$t$" part as follows $$ \frac{dx}{dt} = f(x) g(t) \implies \int \frac{1}{f(x)} \, dx = \int g(t) \, dt $$

In your case we get $$ \frac{dx}{dt} = \underbrace{\left(x^3 \right)}_{\color{blue}{f(x)}} \underbrace{(a)}_{\color{blue}{g(t)}} \color{purple}{\implies} \int \frac{1}{x^3} \, dx = \int a \, dt \color{purple}{\implies}- \frac{1}{2x^2} = at + C^* \color{purple}{\implies} \boxed{x(t) = \frac{1}{\sqrt{-2at + C}}} $$ where $C$ is some arbitrary constant.

Let us consider the following change of variable: $y=x^{-2}$ then \begin{equation}\label{as} x^2 = \frac{1}{y} \ \ \ \Longrightarrow \ \ \ 2\dot{x}x = -\frac{1}{y^2} \dot{y} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \end{equation} let's multiply by 2x the equation $\dot{x} = ax^3$, \begin{equation} 2\dot{x}x = 2ax^4 \ \ \ \ \ \mbox{(here I had an error)} \end{equation} using equation (1) we obtain that \begin{equation} -\frac{1}{y^2}\dot{y} = \frac{2a}{y^2} \ \ \ \Longrightarrow \ \ \ \dot{y} = -2a \end{equation} integrating we obtain \begin{equation} y(t) = -2at + C. \end{equation} Remembering that $y=x^{-2}$ the solution will be as \begin{equation} x(t) = \frac{1}{\sqrt{-2at + C}}. \end{equation}