10. If $a$, $b$, $c$ are the three real roots of the equation
$$
x^{3}-x^{2}-x+m=0
$$

then the minimum value of $m$ is $\qquad$
10. $-\frac{5}{27}$.

From the problem, we have
$$
\begin{array}{l}
x^{3}-x^{2}-x+m=(x-a)(x-b)(x-c) \\
=x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c \text {. } \\
\text { Therefore, }\left\{\begin{array}{l}
a+b+c=1, \\
a b+b c+c a=-1, \\
m=-a b c .
\end{array}\right. \\
\end{array}
$$

From the system of equations, we get
$$
\begin{array}{l}
b c=-1-(a b+c a)=-1-a(b+c) \\
=-1-a(1-a)=a^{2}-a-1, \\
m=-a b c=-a^{3}+a^{2}+a . \\
\text { Also, } a^{2}+b^{2}+c^{2} \\
=(a+b+c)^{2}-2(a b+b c+c a)=3,
\end{array}
$$

Thus, at least one of $a^{2} \leqslant 1, b^{2} \leqslant 1, c^{2} \leqslant 1$ holds.
Without loss of generality, assume $a^{2} \leqslant 1$, i.e., $-1 \leqslant a \leqslant 1$.
Let $m=f(a)=-a^{3}+a^{2}+a$. Then
$$
\begin{array}{c}
f^{\prime}(a)=-3 a^{2}+2 a+1 \\
=-(3 a+1)(a-1) .
\end{array}
$$

Therefore, when $-10$.
Thus, $f(a)$ is a decreasing function on the interval $\left[-1,-\frac{1}{3}\right]$ and an increasing function on the interval $\left[-\frac{1}{3}, 1\right]$.
Hence, $m$ has a minimum value $f\left(-\frac{1}{3}\right)=-\frac{5}{27}$, at which point,
$$
a=-\frac{1}{3}, b=-\frac{1}{3}, c=\frac{5}{3}
$$

or $a=-\frac{1}{3}, b=\frac{5}{3}, c=-\frac{1}{3}$.