Question

The value of ${\omega }^{192}+{\omega }^{194}$ is p. Find $p+\omega$

Answer 1

Since, $1+w+w^2=0$ and $w^3=1$.

Now, $p+w=w^{192}[1+w^2]+w$

$=\left(w^3{)}^{64}\right[1+w^2]+w$

$=\left(1{)}^{64}\right[1+w^2]+w=1+w+w^2=0$