Show 1 + cisx + cis2x = (1-cis3x)/(1-cisx)?

we know
cis x = cos x + i sin x

(cis x)^n = cos ^n x + i sin ^n x = cis nx as per http://en.wikipedia.org/wiki/De_Moivre%2…


so LHS = 1 + cisx + cis2x = 1 + cis x + cis ^2 2x

let cis x = t (not required but makes understanding easier)

LHS = 1 + t + t^2 = (1-t^3)/(1- t) = (1- cis ^3 x)/(1- cis x) =
= ( 1- cis 3x )/(1- cis x) = RHS