= Solution
Take the <zero-boundary Sobolev space> $X=H_0^1(\Omega)$ with norm $\|u\|_X=\|\nabla u\|_2$. The <Poincaré inequality> makes this equivalent to the usual <Sobolev norm>. Define
$$
\boxed{\langle A(u),v\rangle=\int_\Omega\nabla u\cdot\nabla v+\int_\Omega u^3v,\qquad\langle f,v\rangle=\int_\Omega v.}
$$
The <weak formulation> is $u\in X$ and $\langle A(u),v\rangle=\langle f,v\rangle$ for every $v\in X$.
In dimension three the <Sobolev embedding theorem> gives $\|u\|_6\leq C_6\|u\|_X$, so $u^3\in L^2$. The <Cauchy-Schwarz inequality> and <Poincaré inequality> yield
$$
|\langle A(u),v\rangle|\leq\bigl(\|u\|_X+C_PC_6^3\|u\|_X^3\bigr)\|v\|_X,
\qquad|\langle f,v\rangle|\leq|\Omega|^{1/2}C_P\|v\|_X.
$$
Thus $A(u)\in X^*$ for every $u$, and $f\in X^*$. Finally
$$
\begin{aligned}
\langle A(u)-A(w),u-w\rangle
&=\|u-w\|_X^2+\int_\Omega(u-w)^2(u^2+uw+w^2)\\
&\geq\boxed{\|u-w\|_X^2},
\end{aligned}
$$
since $u^2+uw+w^2\geq0$. Hence $A$ is strongly monotone with constant one in the chosen norm.
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