6. Given that $\alpha, \beta$ are two imaginary roots of the equation $2 x^{4}-6 x^{3}+11 x^{2}-6 x-4=0$, then for any integer $p$ not less than 5, we have ( ).
A. $\alpha^{p}-\beta^{p}>(\alpha+\beta)^{p}$
B. $\alpha^{p}+\beta<(\alpha+\beta)^{p}$
C. $\alpha^{p}+\beta^{p}=(\alpha+\beta)^{p}$
D. Any of the above is possible
6. C

From $f(x)=\left(x^{2}-2 x+4\right)\left(2 x^{2}-2 x-1\right)$, we get that $\alpha, \beta$ are the roots of $x^{2}-2 x+4$. They are $1 \pm \sqrt{3} \mathrm{i}$, which can be written as $2\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)$. Therefore, $(\alpha+\beta)^{p}=2^{p}$, and $\alpha^{p}+\beta^{p}=2^{p+1} \cos \frac{p \pi}{3}$. Since $p$ is a prime number greater than 5, $p \equiv -1, 1 (\bmod 6)$. Thus, $\cos \frac{p \pi}{3}=\frac{1}{2}$, and hence $\alpha^{p}+\beta^{p}=(\alpha+\beta)^{p}$.