10. $a, b, m, n$ satisfy: $a m m^{2001}+b n^{2001}=3 ; a m^{20002}+b n^{2002}=7 ; a m^{2003}+$ $b n^{2003}=24 ; a m^{21004}+b m^{2004}=102$. Then the value of $m^{2}(n-1)$ is $\qquad$ .
10.6

From $\left\{\begin{array}{l}a m^{2004}+b n^{2004}=\left(a m^{2003}+b n^{2003}\right)(m+n)-\left(a m^{20012}+b n^{20002}\right) m n \\ a m^{2003}+b m^{2003}=\left(a m^{2002}+b n^{2002}\right)(m+n)-\left(a m^{2001}+b n^{2001}\right) m n\end{array}\right.$
we have $\left\{\begin{array}{l}102=24(m+n)-7 m n \\ 24=7(m+n)-3 m n\end{array}\right.$, which simplifies to $\left\{\begin{array}{l}m+n=6 \\ m n=6\end{array}\right.$. Therefore, $m, n$ are the roots of the equation $x^{2}-6 x+6$
$=0$, so $m^{2}=6 m-6$, and thus $m^{2}(n-1)=(6 m-6)(n-1)=6(m n$
$$
-m-n+1)=6
$$