If $\omega$ is an imaginary cube root of unity and $x = a + b,\ y = a\omega + b\omega^{2},\ \ z = a\omega^{2} + b\omega$, then $x^{2} + y^{2} + z^{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
$x^{2} + y^{2} + z^{2} = (a + b)^{2} + \omega^{2}(a + b\omega)^{2} + \left( a\omega^{2} + b\omega \right)^{2}$
$= a^{2} + b^{2} + 2ab + a^{2}\omega^{2} + b^{2}\omega^{4} + 2ab\omega^{3} + a^{2}\omega^{4} + b^{2}\omega^{2} + 2ab\omega^{3}$
$= a^{2}\left( 1 + \omega + \omega^{2} \right) + b^{2}\left( 1 + \omega + \omega^{2} \right) + 6ab\ \ \ \left\lbrack \because\omega^{4} = \omega \right\rbrack$
$= 6ab\ \ \ \ \ \ \lbrack\because 1 + \omega + \omega^{2} = 0\rbrack$

Q. If $\omega$ is an imaginary cube root of unity and $x = a + b,\ y = a\omega + b\omega^{2},\ \ z = a\omega^{2} + b\omega$, then $x^{2} + y^{2} + z^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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