= Solution
Put $V=\mathbb C^n$. We use <complete reducibility of compact-group representations>, and the highest-weight classification: the multiplicity of $V_\lambda$ in a completely <reducible representation> is the <dimension> of its weight-$\lambda$ subspace killed by all <positive root> vectors. Each <irreducible> summand contributes just its one-dimensional highest-weight line to that subspace.
Symmetrizing and antisymmetrizing the three tensor factors give <invariant subspaces> $\operatorname{Sym}^3V$ and $\Lambda^3V$ of $V^{\otimes3}$. The vectors $e_1^{\otimes3}$ and $e_1\wedge e_2\wedge e_3$ are <highest-weight vectors> of <weights> $(3,0,\ldots,0)$ and $(1,1,1,0,\ldots,0)$ respectively. The unitary version of the <Weyl dimension formula> is
$$
\dim V_\lambda=\prod_{i<j}\frac{\lambda_i-\lambda_j+j-i}{j-i}.
$$
It gives
$$
\dim V_{(3)}=\prod_{j=2}^n\frac{j+2}{j-1}=\binom{n+2}3,
\qquad
\dim V_{(1,1,1)}=\prod_{i=1}^3\frac{n-i+1}{4-i}=\binom n3.
$$
These are exactly the <dimensions> of the third <symmetric power> and <exterior power>, so each is <irreducible> and occurs once in those subspaces.
The <weight> $(2,1,0,\ldots,0)$ in $V^{\otimes3}$ has basis
$$
a=e_1\otimes e_1\otimes e_2,\quad
b=e_1\otimes e_2\otimes e_1,\quad
c=e_2\otimes e_1\otimes e_1.
$$
The only <positive root> vector that acts nontrivially on this <weight space> is $E_{12}$. Acting on the <tensor product> as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes3}$. Consequently the <highest-weight vectors> in this space are precisely
$$
Aa+Bb+Cc\quad\text{with}\quad A+B+C=0.
$$
This is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its <dimension> is
$$
\dim V_{(2,1)}=2\prod_{j=3}^n\frac{j+1}{j-2}
=\frac{n(n^2-1)}3.
$$
The <dimensions> of the summands already found add to
$$
\binom{n+2}3+\binom n3+2\frac{n(n^2-1)}3=n^3.
$$
Thus no other <irreducible> summand can remain. We have proved the full decomposition
$$
\boxed{V^{\otimes3}\cong V_{(3,0,\ldots)}\oplus V_{(1,1,1,0,\ldots)}\oplus V_{(2,1,0,\ldots)}^{\oplus2}.}
$$
There are four summands counted with multiplicity and three distinct <isomorphism> types.
For $n=2$, the exterior cube vanishes. The third <symmetric power> still has <highest weight> $(3,0)$ and <dimension> $4$, while the same two-dimensional highest-vector calculation gives two copies of the <representation> of <weight> $(2,1)$ and <dimension> $2$. It is $\det\otimes V$, since a <determinant twist> shifts the standard <highest weight> $(1,0)$ to $(2,1)$. Hence
$$
\boxed{(\mathbb C^2)^{\otimes3}\cong\operatorname{Sym}^3\mathbb C^2\oplus(\det\otimes\mathbb C^2)^{\oplus2},\qquad8=4+2+2.}
$$
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