The first thing to observe is that the right hand side $u^3-u$ is positive if $u>1$and negative if $u<-1$. Now suppose that at some point $u$ becomes larger than $1$. This would mean the maximum of $u$ is larger than $1$. Let $x\in\Omega$ be such a maximum point. It is obvious that $x$ must be an interior point, since $u=0$ at the boundary. Let us look at $\Delta u$ at $x$. Of course, $\Delta u\leq0$ at $x$. But we know that $u^3-u>0$ at $x$, leading to the conclusion that at $x$, the equation $\Delta u = u^3 - u$ cannot be satisfied, hence $u$ is not a solution to our equation.
I leave the consideration of a minimum with $u<-1$ to you.