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If $x=\omega-\omega^{2}-2,$ then the value of $x^{4}+3 x^{3}+2 x^{2}-11 x-6$ is
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We have, $x=\omega-\omega^{2}-2$ or $x+2=\omega-\omega^{2}$
Squaring, $x^{2}+4 x+4=\omega^{2}+\omega^{4}-2 \omega^{3}$
$=\omega^{2}+\omega^{3} \omega \cdot-2 \omega^{3}=\omega^{2}+\omega-2 \quad\left[\omega^{3}=1\right]$
$=-1-2=-3 \Rightarrow$
$x^{2}+4 x+7=0$
Dividing $x^{4}+3 x^{3}+2 x^{2}-11 x-6$ by $x^{2}+4 x+7$
we get $x^{4}+3 x^{3}+2 x^{2}-11 x-6=\left(x^{2}+4 x+7\right)\left(x^{2}-x-1\right)+1$
$=(0)\left(\mathrm{x}^{2}-\mathrm{x}-1\right)+1=0+1=1$
Squaring, $x^{2}+4 x+4=\omega^{2}+\omega^{4}-2 \omega^{3}$
$=\omega^{2}+\omega^{3} \omega \cdot-2 \omega^{3}=\omega^{2}+\omega-2 \quad\left[\omega^{3}=1\right]$
$=-1-2=-3 \Rightarrow$
$x^{2}+4 x+7=0$
Dividing $x^{4}+3 x^{3}+2 x^{2}-11 x-6$ by $x^{2}+4 x+7$
we get $x^{4}+3 x^{3}+2 x^{2}-11 x-6=\left(x^{2}+4 x+7\right)\left(x^{2}-x-1\right)+1$
$=(0)\left(\mathrm{x}^{2}-\mathrm{x}-1\right)+1=0+1=1$
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