1. ## Cube Roots Of Unity (Quadratic Equations)

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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.

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

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

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

Now just notice that $\displaystyle \Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg a^2)=0$.
This means that $\displaystyle x^2+x+1$ is always equal to 0 and hence $\displaystyle (1+\Omega+\Omega^2)$ = 0, am i correct?

4. It's not always equal to zero, no. For instance it's equal to 3 when $\displaystyle x=1$. However it's equal to 0 when $\displaystyle x$ is a cube root of unity, so $\displaystyle \Omega^2+\Omega+1=0$ is right!

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

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# cube roots of -8 with omega

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