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