10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{2}=5 \text {, }
$$

and $\left(a_{n+1}+a_{n-1}\right)\left(a_{n}^{2}+a_{n} a_{n-1}+a_{n-1}^{2}\right)$
$$
=6\left(a_{n}^{3}-a_{n-1}^{3}\right)\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) \text {. }
$$

Let $c_{n}=\sum_{i=1}^{n}\left[-\frac{i a_{i}}{3^{i}}\right]([x]$ denotes the greatest integer not exceeding the real number $x$), and $b_{m}=-2 m+3(m \in \mathbf{R})$. Find the minimum value of $\left(2 c_{-}+b_{n}\right)^{2}+(n-m)^{2}$.
10. From $a_{n}^{2}+a_{n} a_{n-1}+a_{n-1}^{2}>0$, we know
$$
\begin{array}{l}
a_{n+1}+a_{n}=6\left(a_{n}-a_{n-1}\right) \\
\Rightarrow a_{n+1}=5 a_{n}-6 a_{n-1} \\
\Rightarrow a_{n+1}-2 a_{n}=3\left(a_{n}-2 a_{n-1}\right) \\
\Rightarrow a_{n+1}-2 a_{n}=3^{n-1}\left(a_{2}-2 a_{1}\right)=3^{n} \\
\Rightarrow \frac{a_{n+1}}{3^{n}}-\frac{2 a_{n}}{3 \times 3^{n-1}}=1 \\
\Rightarrow \frac{a_{n+1}}{3^{n}}-3=\frac{2}{3}\left(\frac{a_{n}}{3^{n-1}}-3\right) \\
\Rightarrow \frac{a_{n}}{3^{n-1}}-3=-2 \times\left(\frac{2}{3}\right)^{n-1} \\
\Rightarrow a_{n}=3^{n}-2^{n} .
\end{array}
$$

Then $c_{n}=\sum_{i=1}^{n}\left[\frac{i\left(3^{i}-2^{i}\right)}{3^{i}}\right]=\sum_{i=1}^{n-1} i=\frac{n^{2}-n}{2}$.
Let $P\left(n, 2 c_{n}\right)$ be an integer point on the parabola $y=x^{2}-x$, and $Q\left(m,-b_{m}\right)$ be any point on the line $y=2 x-3$. Then
$$
\left(2 c_{n}+b_{m}\right)^{2}+(n-m)^{2}=|P Q|^{2} \text {. }
$$

The distance from point $P$ to the line $2 x-y-3=0$ is
$$
d=\frac{\left|2 n-2 c_{n}-3\right|}{\sqrt{5}}=\frac{\left|(2 n-3)^{3}-21\right|}{4 \sqrt{5}} \text {. }
$$

When $2 n-3=5$, i.e., $n=4$, $d_{\text {min }}=\frac{\sqrt{5}}{5}$.
Thus, $|P Q|^{2} \geqslant d_{\min }^{2}=\frac{1}{5}$, with equality holding if and only if $n=4$, $c_{n}=6$.
Therefore, the minimum value of the required expression is $\frac{1}{5}$.