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Q)

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

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

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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:
$x=a+b,y=a\omega+b\omega^2,z=a\omega^2+b\omega$
$x+y+z=a+b+a\omega+b\omega^2+a\omega^2+b\omega$
$\qquad\;\;\;\;\;\;=a(1+\omega+\omega^2)+b(1+\omega+\omega^2)$
We know that $(1+\omega+\omega^2)=0$
$\qquad\;\;\;\;\;\;=0$
Step 2:
$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$
We know that $x+y+z=0$
Therefore $x^3+y^3+z^3=3xyz$
We know that $xyz=a^3+b^3$
$\Rightarrow 3(a^3+b^3)$
Therefore $x^3+y^3+z^3=3(a^3+b^3)$
Hence proved.