Question

If $f\left(x\right)=\frac{x^4+x^2+1}{x^2-x+1}$ then find the value of $f\left({\omega }^n\right)$
(where ${}^{\prime }{\omega }^{\prime }$ is the non-real root of the equation $z^3=1$ and ${}^{\prime }{n}^{\prime }$ is a multiple of $3$).

Answer 1

Now, $f\left(x\right)=\frac{x^4+x^2+1}{x^2-x+1}$
$\Rightarrow f\left(x\right)=x^2+x+1$
Now, $f\left({\omega }^n\right)={\omega }^{2n}+{\omega }^n+1=3$
$\because {\omega }^n=1$ when n is a multiple of 3.