= Solution
By continuity on the compact closure, $v$ has a maximum and minimum. If its maximum were greater than one, the boundary hypothesis would force that maximum to occur at an interior point $x_0$. There $\Delta v(x_0)\leq0$, whereas
$$
v(x_0)^3-v(x_0)=v(x_0)\bigl(v(x_0)^2-1\bigr)>0,
$$
a contradiction. If its minimum were less than minus one, it would again be interior, where $\Delta v\geq0$ but $v^3-v<0$. Therefore
$$
\boxed{-1\leq v\leq1\quad\text{on }\overline\Omega,\qquad \sup_\Omega|v|\leq1.}
$$
This is a direct maximum/minimum argument for a <nonlinear elliptic boundary value problem>; treating its nonlinearity as a zeroth-order coefficient of fixed nonpositive sign would not be justified.
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