6. Polynomial sequence: $f_{0}(x)=1, f_{n+1}|x|=\left(x^{2}-1\right) \hat{f}_{n}(x)-2 x(n=0,1,2$, $\cdots$ ) Then the sum of the absolute values of the coefficients of $f_{6}(x)$ is ().
A. 160
B. 230
C. 202
D. 190
6. D

Since $f_{n+1}(x)-\frac{2}{x}=\left(f_{n}(x)-\frac{2}{x}\right)\left(x^{2}+1\right)$,
thus $f_{\pi}(x)=\frac{2}{x}+\left(x^{2}+1\right)^{n}\left(f_{0}(x)-\frac{2}{x}\right)$
$$
=\left(x^{2}+1\right)^{n}-\frac{2}{x}\left[\left(x^{2}+1\right)^{n}-1\right]
$$
$\left(x^{2}+1\right)^{n}$ is the sum of all even powers of $f_{n}(x)$. The coefficients are all positive, and the sum of their absolute values is $(1+1)^{n}=$ $2^{n}-\frac{2}{x}\left[\left(x^{2}+1\right)^{n}-1\right]$ is the sum of all odd powers of $f_{n}(x)$. The coefficients are all negative, and the sum of their absolute values is $\frac{2}{1}$ $\left[(1+1)^{n}-1\right]=2^{n+1}-2$.
Therefore, the sum of the absolute values of all coefficients of $f_{6}(x)$ is $2^{6}+2^{7}-2=190$.