$\begin{aligned}
\Delta & =\left|\begin{array}{llc}
1 & 1 & 1 \
1 & 2 & 4 \
1 & 4 & 10
\end{array}\right|=1(20-16)-1(10-4)+1(4-2) \
& =4-6+2=0
\end{aligned}Forinfinitesolutions\begin{aligned}
& \Delta_x=\Delta_y=\Delta_z=0 \
& m^2-3 x+2=0 \
& m=1,2 \
& \alpha=1, \beta=2 \
& \therefore \sum_{n=1}^{10}\left(n^\alpha+n^\beta\right)=\sum_{n=1}^{10} n^1+\sum_{n=1}^{10} n^2 \
&= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \
&= 55+385 \
&= 440
\end{aligned}$