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=(1-w^3)/(1-w)`

provided that
`w\\eq1`. Any other cube root of unity will make the numerator vanish, so the equality holds.