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Q)

Prove that if $\omega^3=1$,then (ii) $\bigg(\large\frac{-1+i\sqrt 3}{2}\bigg)^5+\bigg(\large\frac{-1-i\sqrt 3}{2}\bigg)^5$$=-1$

This is the second part of the multi-part question Q3

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A)
Toolbox:
  • If $\omega$ is a cube root of unity then $\omega^3=1$ and $1+\omega+\omega^2=0$
Step 1:
$\omega=\large\frac{-1+i\sqrt 3}{2}$ and $\omega^3=1$
$\Rightarrow \omega^2=\large\frac{1}{\omega}=\large\frac{-1-i\sqrt 3}{2(\Large\frac{1}{4}+\frac{3}{4})}$
$\Rightarrow \large\frac{-1-i\sqrt 3}{2}$
Step 2:
LHS=$\omega^5+(\omega^2)^5=\omega^3.\omega^2+(\omega^3)^3\omega$
$\Rightarrow \omega^2+\omega=-1$
Hence proved.
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