6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$
6.2 .

Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega^{2}+a_{2} \omega+a_{3}+a_{4} \omega^{2}+\cdots+ \\
a_{4026}+a_{4027} \omega^{2}+a_{4028} \omega=1 .
\end{array}
$$

Adding the above two equations, we get
$$
\begin{array}{l}
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+ \\
2 a_{405}-a_{4027}-a_{4028}=2 .
\end{array}
$$