Question

Prove that: $\frac{(a+bw+c{w}^2)}{(c+aw+b{w}^2)}=w^2$, where $w$ being cube root of unity.

Answer 1

Since $w$ being cube root of unity then $w^3=1$.

Then,

$\frac{(a+bw+c{w}^2)}{(c+aw+b{w}^2)}$

$=w^2\frac{(a+bw+c{w}^2)}{w^2(c+aw+b{w}^2)}$

$=w^2\frac{(a+bw+c{w}^2)}{(a+bw+c{w}^2)}$ [ Since $w^3=1,w^4=w$.]

$=w^2$.