= Solution
At a primitive cube <root of unity>, $[1]=1$, $[2]=-1$, $[3]=0$. In the degree-three <quantum plane> <module>, put $z_0=x^3$, $z_1=x^2y$, $z_2=xy^2$, $z_3=y^3$. The complete actions are
$$
\begin{array}{c|rrrr}
&z_0&z_1&z_2&z_3\\\hline
K&z_0&qz_1&q^{-1}z_2&z_3\\
E&0&z_0&-z_1&0\\
F&0&-z_2&z_3&0
\end{array}.
$$
Hence $W=\langle z_0,z_3\rangle$ is a two-dimensional trivial <submodule>; every line in $W$ is a <simple module>. The quotient by $W$ is a two-dimensional <simple module>, with distinct $K$ <eigenvalues> $q,q^{-1}$ and nonzero maps connecting its two <weight spaces>.
Let $S$ be a <simple module> that is a <submodule> of this cubic component. If $S\cap W\ne0$, simplicity makes $S\subseteq W$, so $S$ is one of its lines. If $S\cap W=0$, the quotient map injects $S$ into the simple quotient. Its nonzero image is the entire quotient, so $\dim S=2$. Since $K$ is diagonalizable, such an $S$ must contain the unique $q$ <weight space> $\mathbb C z_1$. But $Ez_1=z_0\ne0$, which then belongs to $S\cap W$, a contradiction. Therefore the <cubic-root quantum-plane socle> gives the complete answer:
$$
\boxed{S=\mathbb C(\alpha x^3+\beta y^3),\qquad(\alpha,\beta)\ne(0,0).}
$$
All simple <submodules> lie in the two-dimensional $W$, so their sum cannot equal the four-dimensional cubic component. \b[It is not a direct sum of simple modules.] The obstruction is the nonzero extension to the simple two-dimensional quotient, not a failure of that quotient to be simple.
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