Question

If $x=\omega -{\omega }^2-2,$ then the value of $x^4+3x^3+2x^2-11x-6$ is

Answer 1

$x=\omega -{\omega }^2-2$
$\Rightarrow x=\omega +(\omega +1)-2=2\omega -1$
$(x+1{)}^3=(2\omega {)}^3$
$\Rightarrow x^3+3x^2+3x-7=0$
Now, $x^4+3x^3+2x^2-11x-6$
$=x(x^3+3x^2+3x-7-x-4)-6$
$=x(-x-4)-6$
$=-(x^2+4x+6)$
$=-\left[\right(x+2{)}^2+2]$
$=-\left[\right(2\omega +1{)}^2+2]$
$=-[4{\omega }^2+4\omega +3]$
$=-\left[4\right({\omega }^2+\omega +1)-1]$
$=1$