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If $\omega$ is a complex cube root of unity and $x=\omega^{2}-\omega-2$, then what is the value of $x^{2}+4 x+7 ?$
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Given $x=\omega^{2}-\omega-2$
$\Rightarrow x+2=\omega^{2}-\omega$
On squaring both sides, we get
$(x+2)^{2}=\left(\omega^{2}-\omega\right)^{2}$
$\Rightarrow x^{2}+4 x+4=\omega^{4}+\omega^{2}-2 \omega^{3}$
Add 3 on both side
$\Rightarrow x^{2}+4 x+4+3=\omega+\omega^{2}-2+3\left(\because \omega^{3}=1\right)$
$\Rightarrow x^{2}+4 x+7=1+\omega+\omega^{2}$
$\Rightarrow \quad x^{2}+4 x+7=0\left(\because 1+\omega+\omega^{2}=0\right)$
$\Rightarrow x+2=\omega^{2}-\omega$
On squaring both sides, we get
$(x+2)^{2}=\left(\omega^{2}-\omega\right)^{2}$
$\Rightarrow x^{2}+4 x+4=\omega^{4}+\omega^{2}-2 \omega^{3}$
Add 3 on both side
$\Rightarrow x^{2}+4 x+4+3=\omega+\omega^{2}-2+3\left(\because \omega^{3}=1\right)$
$\Rightarrow x^{2}+4 x+7=1+\omega+\omega^{2}$
$\Rightarrow \quad x^{2}+4 x+7=0\left(\because 1+\omega+\omega^{2}=0\right)$
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