Solution
= Solution
For the <iteration of a map> $f(x)=\lambda x(1-x^2)$, a <fixed point> satisfies $x[1-\lambda+\lambda x^2]=0$. If $0<\lambda<1$, the bracket is strictly positive, so the sole fixed point is $0$. Its <multiplier of a periodic orbit of an iteration> is $f'(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by <fixed point stability for an iteration>.