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If $\omega$ is a non-real cube root of unity and $x=\omega^2-\omega-3$, then the value of $x^4+6 x^3+10 x^2-12 x-19$ is
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The correct answer is:
5
Given,
$\begin{aligned}
& x=\omega^2-\omega-3 \Rightarrow x=\omega^2-\omega-1-2 \\
& x=\omega^2-(\omega+1)-2 \\
& x=\omega^2+\omega^2-2 \\
& x=2 \omega^2-2 \Rightarrow(x+2)^3=8 \omega^6 \\
& x^3+6 x^2+12 x+8=8 \Rightarrow x^3+6 x^2+12 x=0 \\
& \quad x\left(x^2+6 x+12\right)=0 \Rightarrow x^2+6 x+12=0 \\
& x^4+6 x^3+10 x^2-12 x-19 \\
& \left(x^2-2\right)\left(x^2+6 x+12\right)+5 \Rightarrow 0+5=5
\end{aligned}$
$\therefore$ Value of $x^4+6 x^3+10 x^2-12 x-19$ is 5
$\begin{aligned}
& x=\omega^2-\omega-3 \Rightarrow x=\omega^2-\omega-1-2 \\
& x=\omega^2-(\omega+1)-2 \\
& x=\omega^2+\omega^2-2 \\
& x=2 \omega^2-2 \Rightarrow(x+2)^3=8 \omega^6 \\
& x^3+6 x^2+12 x+8=8 \Rightarrow x^3+6 x^2+12 x=0 \\
& \quad x\left(x^2+6 x+12\right)=0 \Rightarrow x^2+6 x+12=0 \\
& x^4+6 x^3+10 x^2-12 x-19 \\
& \left(x^2-2\right)\left(x^2+6 x+12\right)+5 \Rightarrow 0+5=5
\end{aligned}$
$\therefore$ Value of $x^4+6 x^3+10 x^2-12 x-19$ is 5
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