= Solution
The cubic <Taylor expansion> at the double-zero point is
$$
\begin{aligned}
\dot q&=2q-2p-\frac43q^3+2p^2q+pq^2+\frac43p^3+O(5),\\
\dot p&=2q-2p-\frac13q^3-2p^2q+\frac43p^3+O(5).
\end{aligned}
$$
There are no quadratic or quartic terms because the local <vector field> is odd under $(q,p)\mapsto(-q,-p)$. Set $u=p$, $v=q-p$, and use $\tau=2t$. The <linear transformation> is invertible, and the positive time rescaling preserves orbit orientation. With primes denoting $d/d\tau$, the result is
$$
\begin{aligned}
u'&=v-\frac12u^3-\frac32u^2v-\frac12uv^2-\frac16v^3+O(5),\\
v'&=2u^3+\frac32u^2v-uv^2-\frac12v^3+O(5).
\end{aligned}
$$
This supplies the requested cubic functions, with a remainder stronger than $O(4)$.
There is an important <normal form> check. Setting $x=u$, $y=u'$ straightens the first equation, and differentiating gives
$$
x'=y,\qquad y'=2x^3-4xy^2-y^3+O(5).
$$
In particular the <nilpotent cubic damping invariant> is $g_{21}+3f_{30}=3/2-3/2=0$. A nonsingular cubic <near-identity transformation> and rescaling cannot change this vanishing invariant into the nonzero $-x^2y$ coefficient in the supplied later <normal form>. The conclusions for that assumed system are therefore given separately from the independently derived conclusions for the printed original equations.
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