2. For real numbers $a, b, c$ satisfying
$$
a+b+c=0, \, abc=2 \text{.}
$$

then $u=|a|^{3}+|b|^{3}+|c|^{3}$ has the minimum value of
$\qquad$.
2.10.

From the problem, we know that $a$, $b$, and $c$ must be one positive and two negative.
Without loss of generality, let $a>0$, $b<0$, and $c<0$.
Since $b+c=-a$ and $bc=\frac{2}{a}$, it follows that $b$ and $c$ are the two negative roots of the equation $x^{2}+a x+\frac{2}{a}=0$. Therefore, we have
$$
\Delta=a^{2}-\frac{8}{a} \geqslant 0 \text {. }
$$

Solving this, we get $a^{3} \geqslant 8$.
Thus, $u=a^{3}-b^{3}-c^{3}$
$$
\begin{array}{l}
=a^{3}-(b+c)\left[(b+c)^{2}-3 b c\right] \\
=a^{3}+a\left(a^{2}-\frac{6}{a}\right)=2 a^{3}-6 \\
\geqslant 2 \times 8-6=10 .
\end{array}
$$

Equality holds if and only if $a^{3}=8$, i.e., $a=2$.
In this case, $b=c=-1$.
Therefore, the minimum value of $u$ is 10.