[JEE Advanced 1978] if x a b y a omega b omega 2 and z a omega
If $x=a+b$, $y=a\omega+b\omega^2$ and $z=a\omega^2+b\omega$ ($\omega$ a complex cube root of unity), show that $xyz=a^3+b^3$.
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· 2d ago
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Answer: $xyz=a^3+b^3$.
$(a\omega+b\omega^2)(a\omega^2+b\omega)=a^2\omega^3+ab(\omega^2+\omega^4)+b^2\omega^3=a^2-ab+b^2$; times $(a+b)$ gives $a^3+b^3$.
JEE Advanced 1978 · Complex Numbers — verified solution by the Vidaara Team.
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