10. Let $n$ be a composite positive integer, $f(n)$ be the sum of the smallest three positive divisors of $n$, and $g(n)$ be the sum of the largest two positive divisors of $n$. If $g(n)=f^{3}(n)$, then all possible values of $n$ are $\qquad$
10. 144.

Let $n$ be a positive integer that satisfies the condition.
Obviously, the smallest and largest positive divisors of $n$ are 1 and $n$, respectively. Let $p$ be the smallest prime divisor of $n$, then the second smallest and second largest positive divisors of $n$ are $p$ and $\frac{n}{p}$, respectively.

For the third smallest positive divisor of $n$, there are two cases:
(1) If the third smallest positive divisor is $p^{2}$, then
$$
p^{2} \mid n, f(n)=1+p+p^{2}, g(n)=n+\frac{n}{p} \text{. }
$$

From $p^{2} \mid n$, we know that $g(n)=n+\frac{n}{p}$ is a multiple of $p$, so
$$
\begin{array}{l}
g(n) \equiv 0(\bmod p) . \\
\text { Also, } f(n)=1+p+p^{2} \equiv 1(\bmod p), \\
f^{3}(n) \equiv 1(\bmod p),
\end{array}
$$

Thus, $g(n) \neq f^{3}(n)$, which contradicts the condition.
(2) If the third smallest positive divisor is a prime $q$ $(q>p)$, then
$$
\begin{array}{l}
f(n)=1+p+q \equiv 1+p(\bmod q), \\
f^{3}(n) \equiv(1+p)^{3}(\bmod q) .
\end{array}
$$

From $p q \mid n$, we know that $g(n)=n+\frac{n}{p}$ is a multiple of $q$, so
$$
g(n) \equiv 0(\bmod q) \text{. }
$$

Thus, $(1+p)^{3} \equiv 0(\bmod q)$.
Since $q$ is a prime, we have $q \mid(1+p)$.
Therefore, $p<q \leqslant 1+p$, and the only primes $p$ and $q$ that satisfy this are $p=2, q=3$. In this case,
$$
f(n)=6, g(n)=\frac{3 n}{2} \text{. }
$$

From $g(n)=f^{3}(n)$, we get
$$
\frac{3 n}{2}=6^{3} \Rightarrow n=144 \text{. }
$$

Upon verification, $n=144$ meets the requirements.
Therefore, the only possible value of $n$ is $n=144$.