Q. If $x = a + b,\ y + a\alpha + b\beta\ and\ \mathcal{z =}a\beta + b\alpha,\ where\ \alpha\ and\ \beta$ are complex cube roots of unity, then $xy\mathcal{z}$ is equal to
A) $a^{2} + b^{2}$
B) $a^{3} + b^{3}$
C) $a^{3}b^{3}$
D) $a^{3} - b^{3}$
Explanation
(b)
$x = a + b,\ y = a\alpha + b\beta\ \ and\ \mathcal{\ z =}a\beta + b\alpha\$
Now, $xy\mathcal{z =}(a + b)\left( a\omega + b\omega^{2} \right)(a\omega^{2} + b\omega)$,
Where $\alpha = \omega\ \ and\ \ \beta = \omega^{2}$
$\therefore\ xy\mathcal{z =}(a + b)(a^{2} + ab\omega^{2} + ab\omega + b^{2})$,
$= (a + b)\left( a^{2} - ab + b^{2} \right) = a^{3} + b^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.