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If $2 x=-1+\sqrt{3} i$, then the value of $\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}=$
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We have,
$2 x=-1+\sqrt{3} i \Rightarrow x=\frac{-1+\sqrt{3} i}{2}=\omega$
So, $\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}$
$=\left(1-\omega^{2}+\omega\right)^{6}-\left(1-\omega-\omega^{2}\right)^{6}$
$=\left(-2 \omega^{2}\right)^{6}-(-\omega-\omega)^{6}=2^{6} \omega^{12}-2^{6}()^{6}$
$=2^{6}-2^{6}=0 \quad\left[\because \omega^{3}=1\right]$
$2 x=-1+\sqrt{3} i \Rightarrow x=\frac{-1+\sqrt{3} i}{2}=\omega$
So, $\left(1-x^{2}+x\right)^{6}-\left(1-x+x^{2}\right)^{6}$
$=\left(1-\omega^{2}+\omega\right)^{6}-\left(1-\omega-\omega^{2}\right)^{6}$
$=\left(-2 \omega^{2}\right)^{6}-(-\omega-\omega)^{6}=2^{6} \omega^{12}-2^{6}()^{6}$
$=2^{6}-2^{6}=0 \quad\left[\because \omega^{3}=1\right]$
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