(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 a0.

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- Aug 31st 2009, 11:21 PMsaberteethCube 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 PMBruno 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 $\displaystyle \Omega^2+\Omega^3+\Omega^4=\Omega^2(1+\Omega+\Omeg a^2)=0$. - Sep 1st 2009, 01:30 AMsaberteeth
- Sep 1st 2009, 07:42 AMBruno 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! (Rock)

- Sep 1st 2009, 06:55 PMsaberteeth