Question

X=a+b y=aw+bw^2 z=aw+bw^2 omega is cube root of unity calculate xyz

Answer 1

Answer:

$\bf\;xyx=a^3+b^3$

Step-by-step explanation:

Concept:

$\text{If \omega is a cube root of unity, then}\\\\\bf{\omega}^3=1\;\;and\;\;1+\omega+{\omega}^2=0$

Given:

$x=a+b$

$y=a\omega+b{\omega}^2$

$z=a{\omega}^2+b\omega$

$xyx=(a+b)(a\omega+b{\omega}^2)(a{\omega}^2+b\omega)$

$xyx=(a+b)(a^2{\omega}^3+ab{\omega}^2+ab{\omega}^4+b^2{\omega}^3)$

Using $\boxed{\bf{\omega}^3=1}$

$xyx=(a+b)(a^2(1)+ab({\omega}^2+{\omega})+b^2(1))$

Using $\boxed{\bf\;1+{\omega}+{\omega}^2=0\implies\;\;{\omega}+{\omega}^2=-1}$

$xyx=(a+b)(a^2+ab(-1)+b^2)$

$xyx=(a+b)(a^2-ab+b^2)$

$\implies\boxed{\bf\;xyx=a^3+b^3}$

Answer 2

Answer:

yes the above and is correct

Step-by-step explanation:

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