Question

Prove that w^49+w^101+w^150=0,where w is a complex cube root of unity

Answer 1

Because $w$ is a cube root of unity, you have $w^3=1$. So

$w^{49}=w(w^3)^{16}=w$
$w^{101}=w^2(w^3)^{33}=w^2$
$w^{150}=(w^3)^{50}=1$

and so

$w^{49}+w^{101}+w^{150}=1+w+w^2=\dfrac{1-w^3}{1-w}$

provided that $w\neq1$. Any other cube root of unity will make the numerator vanish, so the equality holds.