# Cube Roots Of Unity (Quadratic Equations)

• Aug 31st 2009, 11:21 PM
saberteeth
Cube Roots Of Unity (Quadratic Equations)
(Hi),

If 1, Ω, Ω^2 are the cube roots of unity, find the value of:

1) Ω^2 + Ω^3 + Ω^4:

Ans:

Since Ω^3 = 1

Ω^2 + 1 + Ω

So, Ω^2 + Ω^3 + Ω^4 = 0

I did not understand how the final answer result to be a 0.
• Aug 31st 2009, 11:39 PM
Bruno J.
Cube roots of unity are the roots of $x^3-1$. Note that $x^3-1 = (x-1)(x^2+x+1)$. Since $\Omega$ is not a root of $x-1$ it must be a root of $x^2+x+1$.

Now just notice that $\Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg a^2)=0$.
• Sep 1st 2009, 01:30 AM
saberteeth
Quote:

Originally Posted by Bruno J.
Cube roots of unity are the roots of $x^3-1$. Note that $x^3-1 = (x-1)(x^2+x+1)$. Since $\Omega$ is not a root of $x-1$ it must be a root of $x^2+x+1$.

Now just notice that $\Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg a^2)=0$.

This means that $x^2+x+1$ is always equal to 0 and hence $(1+\Omega+\Omega^2)$ = 0, am i correct?
• Sep 1st 2009, 07:42 AM
Bruno J.
It's not always equal to zero, no. For instance it's equal to 3 when $x=1$. However it's equal to 0 when $x$ is a cube root of unity, so $\Omega^2+\Omega+1=0$ is right! (Rock)
• Sep 1st 2009, 06:55 PM
saberteeth
Quote:

Originally Posted by Bruno J.
It's not always equal to zero, no. For instance it's equal to 3 when $x=1$. However it's equal to 0 when $x$ is a cube root of unity, so $\Omega^2+\Omega+1=0$ is right! (Rock)

I've understood it partially. I think i am unclear with the concept of "cube root of unity" If you can explain what it actually means, i would be grateful.(Nod)