Question

If $\omega$ is 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 $x^{4}+6 x^{3}+10 x^{2}-12 x-19$ is

• A. 5
• B. 7
• C. 12
• D. -19

Answer 1

Answer: (A)

Given
$x=\omega^{2}-\omega-3 $
$\Rightarrow x=\omega^{2}-\omega-1-2 $
$x=\omega^{2}-(\omega+1)-2$
$x=\omega^{2}+\omega^{2}-2$
$\left[\because 1+\omega+\omega^{2}=0\right] $
$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 $
$x\left(x^{2}+6 x+12\right)=0$
$ \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$