Question

If $x=\omega -{\omega }^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is (where $\omega$ is cube root of unity)

Answer 1

Given that $x=\omega -{\omega }^2-2$
$\Rightarrow x+2=\omega -{\omega }^2$
On squaring & rearranging, we get $x^2+4x+7=0$ ....(1)
dividing $x^4+3x^3+2x^2-11x-6$ with $x^2+4x+7$, we get $x^4+3x^3+2x^2-11x-6=(x^2+4x+7)(x^2-x-1)+1$
Therefore using (1), we get $x^4+3x^3+2x^2-11x-6=1$

Ans: 1