1. Tardigrade
  2. Question
  3. Mathematics
  4. Let x =α+β, y =α ω+β ω2, z =α ω2+β ω, ω is an imaginary cube root of unity. Product of xyz is.

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Let $x =\alpha+\beta, y =\alpha \omega+\beta \omega^{2}, z =\alpha \omega^{2}+\beta \omega, \omega$ is an imaginary cube root of unity. Product of $ xyz $ is.

BITSATBITSAT 2005

Solution:

Given, $x=\alpha+\beta, y=\alpha \omega+\beta \omega^{2}$ and $z=\alpha \omega^{2}+\beta \omega$
Also, $\omega^{3}=1$
Now, $x y z=(\alpha+\beta)\left(\alpha \omega+\beta \omega^{2}\right)\left(\alpha \omega^{2}+\beta \omega\right)$
$x y z=\left[\alpha^{2} \omega+\alpha \beta \omega^{2}+\alpha \beta \omega+\beta^{2} \omega^{2}\right]\left[\alpha \omega^{2}+\beta \omega\right]$
$x y z=\alpha^{3} \omega^{3}+\alpha^{2} \beta \omega^{2}+\alpha^{2} \beta \omega^{4}+\alpha \beta^{2} \omega^{3}+\alpha^{2} \beta \omega^{3}+\alpha \beta^{2} \omega^{2}+\alpha \beta^{2} \omega^{4}+\beta^{3} \omega^{3}$
$x y z=\alpha^{3}+\alpha^{2} \beta \omega^{2}+\alpha^{2} \beta \omega+\alpha \beta^{2}+\alpha^{2} \beta+\alpha \beta^{2} \omega^{2}+\alpha \beta^{2} \omega+\beta^{3}$
$\left[\because \omega^{4}=\omega\right]$
$x y z=\alpha^{3}+\beta^{3}+\alpha^{2} \beta\left(1+\omega+\omega^{2}\right)+\alpha \beta^{2}\left(1+\omega+\omega^{2}\right)$
$x y z=\alpha^{3}+\beta^{3}+0+0$
$\left[\because 1+\omega+\omega^{2}=0\right]$
$x y z=\alpha^{3}+\beta^{3}$