Question

If $\omega$ is non-real cube root of unity and $x={\omega }^2-\omega -3$, then the value of $x^4+6x^3+10x^2-12x-19$ is $x^4+6x^3+10x^2-12x-19$ is

• A. 5
• B. 7
• C. 12
• D. -19

Answer 1

Answer: (A)

Given
$x={\omega }^2-\omega -3$
$\Rightarrow x={\omega }^2-\omega -1-2$
$x={\omega }^2-(\omega +1)-2$
$x={\omega }^2+{\omega }^2-2$
$\left[\because 1+\omega +{\omega }^2=0\right]$
$x=2{\omega }^2-2$
$\Rightarrow (x+2{)}^3=8{\omega }^6$
$x^3+6x^2+12x+8=8$
$\Rightarrow x^3+6x^2+12x=0$
$x\left(x^2+6x+12\right)=0$
$\Rightarrow x^2+6x+12=0$
$x^4+6x^3+10x^2-12x-19$
$\left(x^2-2\right)\left(x^2+6x+12\right)+5$
$\Rightarrow 0+5=5$