If 2x = -1 + sqrt{3}i,then the value of (1 - | vedclass

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(D) Given,$2x = -1 + i\sqrt{3}$.$x = \frac{-1 + i\sqrt{3}}{2} = \omega$,where $\omega$ is a complex cube root of unity.We know that $1 + \omega + \ome

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(D) Given,$2x = -1 + i\sqrt{3}$.$x = \frac{-1 + i\sqrt{3}}{2} = \omega$,where $\omega$ is a complex cube root of unity.We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$.Substituting $x = \omega$ into the expression:$(1 - \omega^2 + \omega)^6 - (1 - \omega + \omega^2)^6$$= (-\omega^2 - \omega^2)^6 - (-\omega - \omega)^6$$= (-2\omega^2)^6 - (-2\omega)^6$$= 2^6 \cdot \omega^{12} - 2^6 \cdot \omega^6$Since $\omega^3 = 1$,we have $\omega^6 = 1$ and $\omega^{12} = 1$.$= 64(1) - 64(1) = 0$.

If $2x = -1 + \sqrt{3}i$,then the value of $(1 - x^2 + x)^6 - (1 - x + x^2)^6$ is

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