= Solution
With $x=v+w$ and $y=v-w$, the equations at $r=1$ become
$$
\dot v=\frac12\{a(v+w)^3-z(v-w)\},
$$
$$
\dot w=-2w+\frac12\{a(v+w)^3+z(v-w)\},
\qquad
\dot z=-z+v^2-w^2.
$$
The symmetry $(v,w,z)\mapsto(-v,-w,z)$ suggests
$$
w_c(v)=c_3v^3+O(v^5),\qquad
z_c(v)=d_2v^2+d_4v^4+O(v^6).
$$
Substitute these into the invariance equations
$$
w_c'(v)\dot v=\dot w,\qquad
z_c'(v)\dot v=\dot z.
$$
Comparison at orders $v^2,v^3,v^4$ gives
$$
d_2=1,\qquad c_3=\frac{a+1}{4},\qquad d_4=1-a.
$$
Thus the <center manifold of the 2024 Cambridge cubic system> is
$$
\boxed{w_c(v)=\frac{a+1}{4}v^3+O(v^5),\qquad
z_c(v)=v^2+(1-a)v^4+O(v^6)}.
$$
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