10. Given that $n$ is a natural number, the polynomial $P(x)=\sum_{h=0}^{n} \mathrm{C}_{n}^{h} x^{n-h} \cdot(x-1)^{h}$ can be expanded into an ascending power series of $x$ as $a_{0}+a_{1} x+a_{2} x^{2}+$ $\cdots+a_{n} x^{n}$. Then $\left|a_{0}\right|+\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=$ $\qquad$
10. $3^{n}$.
$$
\begin{array}{l}
P(x)=(x+x-1)^{n}=(2 x-1)^{n} \\
=\sum_{k=0}^{n} \mathrm{C}_{n}^{k}(2 x)^{k}(-1)^{n-k},
\end{array}
$$

Therefore, $\sum_{k=0}^{n}\left|a_{k}\right|=\sum_{k=0}^{n} C_{n}^{k}(2)^{k}=3^{n}$.