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If $x=a+b$,$y=a\omega+b\omega^2,z=a\omega^2+b\omega$ show that (i) $xyz=a^3+b^3$ where $\omega$ is the complex cube root of unity.

This is the first part of the multi-part question Q2

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1 Answer

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  • If $\omega$ is a cube root of unity then $\omega^3=1$ and $1+\omega+\omega^2=0$
Step 1:
We know that $(1+\omega+\omega^2)=0$
Step 2:
We know that $(1+\omega+\omega^2)=0$
$\Rightarrow \omega+\omega^2=-1$
Hence proved.
answered Jun 12, 2013 by sreemathi.v

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