Answer to Question #117202 in Algebra for Asubonteng Isaac Adjei

Question

Given that w denotes either one of the non-real roots of the equation z3 = 1, show that (a) 1 +w+w2 =0, and 3 (b) the other non-real root is w2. Show that the non-real roots of the equation 1 −u u can be expressed in the form Aw and Bw2, where A and B are real numbers, find A and B.

Answer 1

Given equation,

$z^3=1$

or, $z^3-1=(z-1)(z^2+1+z)=0$

or,$z=1 \ and \ z^2+z+1=0$

Since, z=$\omega$ is a root of the equation, then, putting the value of z=$\omega$ in the second equation, we get,

$\omega^2+\omega+1=0$ (hence proved)

Now, in the equation,$z^2+z+1=0$ , sum of the two roots $=\frac{-b}{a}= -1. \$ Since, one root is z=$\omega$, then the other root will be z= -1-$\omega$ . Now from the above relation ($\omega^2+\omega+1=0)$ we get that the other root of the equation z=-1-$\omega$ can be expressed as z=$\omega^2$. (proved).

Next, the other equation is $1-u^2=0$ . Root of this equation are $u=\frac{+}{-}1$. Hence, the roots of this equation can be written as $u=\frac{+}{-}\omega\frac{+}{-}\omega^2.$ Hence, the values of A and B :

A=$\frac{+}{-}1 \ and \ B=\frac{+}{-}1.$