Given, $2 x=-1+i \sqrt{3}$
$\Rightarrow \quad x=\frac{-1+i \sqrt{3}}{2}=w$
which is one of the cube root of unity.
$\therefore \quad\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}$
$\begin{aligned}=\left(1-\omega^{2}+\omega\right)^{6}-\left(1-\omega+\omega^{2}\right)^{6} \\(&\left.\because 1+\omega+\omega^{2}=0\right) \end{aligned}$
$=\left(-\omega^{2}-\omega^{2}\right)^{6}-(-\omega-\omega)^{6}$
$=\left(-2 \omega^{2}\right)^{6}-(-2 \omega)^{6}$
$=2^{6}\left\{\omega^{12}-\omega^{6}\right\}$
$=2^{6}(1-1) \quad\left(\because \omega^{3}=1\right)$
$=2^{6} \cdot 0=0$